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Like any moving object, the motion of an object in free fall can be described by four kinematic equations. The kinematic equations that describe any object's motion are:

The symbols in the above equation have a specific meaning: the symbol d stands for the displacement ; the symbol t stands for the time ; the symbol a stands for the acceleration of the object; the symbol v i stands for the initial velocity value; and the symbol v f stands for the final velocity .

Applying Free Fall Concepts to Problem-Solving

There are a few conceptual characteristics of free fall motion that will be of value when using the equations to analyze free fall motion. These concepts are described as follows:

• An object in free fall experiences an acceleration of -9.8 m/s/s. (The - sign indicates a downward acceleration.) Whether explicitly stated or not, the value of the acceleration in the kinematic equations is -9.8 m/s/s for any freely falling object.
• If an object is merely dropped (as opposed to being thrown) from an elevated height, then the initial velocity of the object is 0 m/s.
• If an object is projected upwards in a perfectly vertical direction, then it will slow down as it rises upward. The instant at which it reaches the peak of its trajectory, its velocity is 0 m/s. This value can be used as one of the motion parameters in the kinematic equations; for example, the final velocity ( v f ) after traveling to the peak would be assigned a value of 0 m/s.
• If an object is projected upwards in a perfectly vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity that it has when it returns to the same height. That is, a ball projected vertically with an upward velocity of +30 m/s will have a downward velocity of -30 m/s when it returns to the same height.

These four principles and the four kinematic equations can be combined to solve problems involving the motion of free falling objects. The two examples below illustrate application of free fall principles to kinematic problem-solving. In each example, the problem solving strategy that was introduced earlier in this lesson will be utilized.  

Example Problem A

Luke Autbeloe drops a pile of roof shingles from the top of a roof located 8.52 meters above the ground. Determine the time required for the shingles to reach the ground.

The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 8.52 meters. The displacement ( d ) of the shingles is -8.52 m. (The - sign indicates that the displacement is downward). The remaining information must be extracted from the problem statement based upon your understanding of the above principles . For example, the v i value can be inferred to be 0 m/s since the shingles are dropped (released from rest; see note above ). And the acceleration ( a ) of the shingles can be inferred to be -9.8 m/s 2 since the shingles are free-falling ( see note above ). (Always pay careful attention to the + and - signs for the given quantities.) The next step of the solution involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the time of fall. So t is the unknown quantity. The results of the first three steps are shown in the table below.

The next step involves identifying a kinematic equation that allows you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are d , v i , a , and t . Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top left contains all four variables.

Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below.

-8.52 m = (0 m/s) • (t) + ½ • (-9.8 m/s 2 ) • (t) 2

-8.52 m = (0 m) *(t) + (-4.9 m/s 2 ) • (t) 2

-8.52 m = (-4.9 m/s 2 ) • (t) 2

(-8.52 m)/(-4.9 m/s 2 ) = t 2

1.739 s 2 = t 2

The solution above reveals that the shingles will fall for a time of 1.32 seconds before hitting the ground. (Note that this value is rounded to the third digit.)

The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The shingles are falling a distance of approximately 10 yards (1 meter is pretty close to 1 yard); it seems that an answer between 1 and 2 seconds would be highly reasonable. The calculated time easily falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for time and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is!  

Example Problem B

Rex Things throws his mother's crystal vase vertically upwards with an initial velocity of 26.2 m/s. Determine the height to which the vase will rise above its initial height.

Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 26.2 m/s. The initial velocity ( v i ) of the vase is +26.2 m/s. (The + sign indicates that the initial velocity is an upwards velocity). The remaining information must be extracted from the problem statement based upon your understanding of the above principles . Note that the v f value can be inferred to be 0 m/s since the final state of the vase is the peak of its trajectory ( see note above ). The acceleration ( a ) of the vase is -9.8 m/s 2 ( see note above ). The next step involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the vase (the height to which it rises above its starting height). So d is the unknown information. The results of the first three steps are shown in the table below.

The next step involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are v i , v f , a , and d . An inspection of the four equations above reveals that the equation on the top right contains all four variables.

v f 2 = v i 2 + 2 • a • d

(0 m/s) 2 = (26.2 m/s) 2 + 2 •(-9.8m/s 2 ) •d

0 m 2 /s 2 = 686.44 m 2 /s 2 + (-19.6 m/s 2 ) •d

(-19.6 m/s 2 ) • d = 0 m 2 /s 2 -686.44 m 2 /s 2

(-19.6 m/s 2 ) • d = -686.44 m 2 /s 2

d = (-686.44 m 2 /s 2 )/ (-19.6 m/s 2 )

The solution above reveals that the vase will travel upwards for a displacement of 35.0 meters before reaching its peak. (Note that this value is rounded to the third digit.)

The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The vase is thrown with a speed of approximately 50 mi/hr (merely approximate 1 m/s to be equivalent to 2 mi/hr). Such a throw will never make it further than one football field in height (approximately 100 m), yet will surely make it past the 10-yard line (approximately 10 meters). The calculated answer certainly falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is!

Kinematic equations provide a useful means of determining the value of an unknown motion parameter if three motion parameters are known. In the case of a free-fall motion, the acceleration is often known. And in many cases, another motion parameter can be inferred through a solid knowledge of some basic kinematic principles . The next part of Lesson 6 provides a wealth of practice problems with answers and solutions.  

Free Fall Motion: Explanation, Review, and Examples

• The Albert Team
• Last Updated On: February 16, 2023

Free fall and projectile motion describe objects that are moving through the air and acted on only by gravity. In this post, we will describe this type of motion using both graphs and kinematic equations. Since projectile motion involves two dimensions, these problems can be complex. We will explain many examples so you can see how to solve different types of projectile motion. 

What We Review

An object that is moving under only the influence of gravity is in free fall. In order for an object to be in free fall, wind and air resistance must be ignored. On Earth, all objects in free fall accelerate downward at the rate of gravity or 9.81\text{ m/s}^2 .

Applying Free Fall to Kinematic Equations

When analyzing free fall motion, we can apply the same kinematic equations as we did for motion on the ground. We can then use these equations to determine properties such as distance, time, and velocity. 

How to Find Distance Fallen for an Object in Free Fall

If an object is in free fall, we can use kinematic equations to find the distance it falls during a certain time. You will typically use the following kinematic equation to calculate the distance fallen:

In order to use this equation, you need to know the initial velocity of the object and the time of flight. Remember that the acceleration of a free falling object is always equal to the acceleration due to gravity, 9.81\text{ m/s}^2 . 

Many free fall physics problems will include scenarios where objects are dropped from rest. In this case, the initial velocity is zero and the first term of the kinematic equation above will cancel out. 

If the time is not known, another method for calculating the distance fallen is to use the following kinematic equation:

In this case, you must know the final velocity v_f of the object. Then, you can solve the equation for the distance d .

How to Find Time for an Object in Free Fall

The amount of time an object is in free fall will depend on its velocity and the distance it falls. Similar to distance, there are two equations you can use to find the time, depending on what you know. 

If you know the initial and final velocity of the object, then the simplest way to calculate time is using the kinematic equation:

This equation can be solved for time. Then, you’ll only need to substitute the values for the velocities and the acceleration due to gravity.

Another method to find time if you do not know the object’s final velocity is to use the equation:

Note that in this equation there are two terms that include the time t . Unless the initial velocity is zero, this can make it more challenging to solve this equation for time. If using this equation, you may need to use the quadratic formula to solve for time.

How to Find Final Velocity for an Object in Free Fall

The final velocity of an object in free fall depends on the amount of time it falls. Due to the acceleration of gravity, the velocity will increase every second by 9.81\text{ m/s} . The final velocity can be calculated using the equation:

If you do not know the amount of time the object is falling, another method for calculating the final velocity is using the kinematic equation: 

This equation requires that you instead know the distance that the object falls. If you are using this equation to find the final velocity, remember that the final velocity is squared in this equation. That means you will need to take a square root as your final step to solve for the final velocity. 

Examples of Free Fall

In this next section, we’ll apply the methods you just learned to solve some problems about free fall motion.

Example 1: How to Find the Distance for an Object Dropped from Rest

For example, an object is dropped from rest from the top of a tall building. It hits the ground 5\text{ s} after it is dropped. What is the height of the building? 

In this scenario, we know that the object’s initial velocity is zero because it was dropped from rest. We also know that the acceleration is 9.81\text{ m/s}^2 . This problem is asking us to find the distance the object falls. This will be equal to the height of the building.

Based on this information, we can use the following kinematic equation to find the distance:

Substituting the given values produces:

Therefore, the height of the building is about 123\text{ m} .

Example 2: How to Find the Final Velocity for an Object with Initial Velocity

In another example, an object in free fall has an initial, downward velocity of 2\text{ m/s} and falls a distance of 45\text{ m} . What is the object’s final velocity? 

In this scenario, we are given the object’s initial velocity, v_i and the distance d . We also know that the acceleration is 9.81\text{ m/s}^2 . Based on this information, we can use the following kinematic equation to find the final velocity:

Since the initial velocity is in the same direction as the acceleration (downward) we can use the same sign for both values.

Our last step is to eliminate the square by taking the square root:

Therefore, the final velocity of the object is about 30\text{ m/s} .

Motion Graphs for Objects in Free Fall

In addition to using physics equations, we can also represent free fall motion with motion graphs. Position-time graphs, velocity-time graphs, and acceleration-time graphs can tell us a lot about the object’s motion over time. Want a more in-depth review of motion graphs? Check out this blog post !

Position-Time Graph for an Object in Free Fall

In terms of position, many objects in free fall start at a high position, or height off the ground, and move downward. Objects in free fall accelerate due to gravity. Therefore, the position-time graph for free fall motion must be curved. This means that objects in free fall start with a slow velocity and gradually speed up which is represented by the steep downward curve of the graph. 

Velocity-Time Graph for an Object in Free Fall

As an object falls, its velocity increases due to the acceleration of gravity. This means that the velocity starts slow and steadily increases in the downward direction. The graph below shows the velocity-time for an object in free fall:

Note that the slope of this graph is constant and represents the acceleration due to gravity, or -9.81\text{ m/s}^2 .

Acceleration-Time Graph for an Object in Free Fall

Free fall acceleration is constant. Throughout the entire time that an object is falling, it is accelerating at a rate equal to the acceleration due to gravity, -9.81\text{ m/s}^2 . As shown in the graph below, the acceleration-time graph is a constant negative line. 

Projectile Motion

A projectile is an object that is launched or thrown into the air and then only influenced by gravity. Projectile motion has many similarities to free fall motion, however, projectiles may also travel a horizontal distance in addition to falling vertically down. 

Examples of Projectile Motion

The exact trajectory, or path, a projectile will take depends on how it is launched. However, all projectiles follow a curved trajectory such as in the image shown below:

If you play or watch sports, you likely have already observed projectile motion. Projectile motion describes the arc of a basketball in a free throw, a fly ball in baseball, or a volleyball bumped over the net. 

Horizontal Component of Velocity

To analyze projectile motion, we must separate the motion into horizontal and vertical components. The horizontal component of a projectile’s velocity is independent of the vertical component of velocity. Since gravity acts vertically, there are no horizontal forces acting on projectiles. This means that the horizontal component of a projectile’s velocity remains constant throughout the entire flight. 

Example: Finding the Horizontal Component

For example, a projectile is launched from the ground with an initial speed of 8\text{ m/s} at a 60^{\circ} angle. What is the horizontal component of the projectile’s velocity?

We will need to use trig identities to determine the components of the velocity. We can visualize the components as a triangle where the hypotenuse is the initial velocity and the sides represent the horizontal, v_{ix} , and vertical, v_{iy} , components of the velocity.

Cosine is defined as the adjacent side of the triangle divided by the hypotenuse. Since the horizontal component is adjacent to the angle, we can use cosine to find the horizontal component of velocity:

Therefore, the horizontal component of the initial velocity is 4\text{ m/s} .

Need to review your trig identities? Try out this resource from Khan Academy .

Vertical Component of Velocity

The vertical component of a projectile’s velocity will be influenced by gravity, which acts vertically on the object causing it to accelerate downward. Therefore, the vertical component of velocity will change throughout the projectile’s flight. We can calculate the vertical component of velocity at a particular time in a method similar to calculating the horizontal component. 

Example: Finding the Vertical Component

In the same example, a projectile is launched from the ground with an initial speed of 8\text{ m/s} at a 60^{\circ} angle. What is the vertical component of the projectile’s velocity?

As we visualize the velocity components, we are solving this time for the opposite side of the triangle. Sine is defined as the opposite side of the triangle divided by the hypotenuse. Therefore, the initial vertical velocity is:

Solving Projectile Motion Questions

Let’s apply what we’ve learned to some examples of projectile motion!

Example 1: Finding the Range of a Projectile

In this example, a projectile is fired horizontally with a speed of 5\text{ m/s} from a cliff with a height of 60\text{ m} . How far from the base of the cliff will the projectile land? 

In this scenario, we are given the initial horizontal velocity v_{ix}=5\text{ m/s} and the vertical change in position d_y=-60\text{ m} . Since the projectile is launched horizontally, the initial vertical velocity, v_{iy} , is zero. We also always know in projectile motion that the vertical acceleration is a_y=-9.81\text{ m/s}^2 and the horizontal acceleration, a_x , is zero.

This problem is asking us to find the horizontal displacement, or d_x . This is also referred to as the range . We can use the following kinematic equation to find the projectile’s final horizontal position:

Since the horizontal acceleration of a projectile is zero, this equation can be simplified to:

Before we can solve this equation, we must first determine the time of the projectile’s flight. We can actually use this same equation in the vertical direction to solve for time:

Since the initial vertical velocity is zero, this equation can be simplified to:

Solving for t :

Substituting the given values:

Now we can use this time to calculate the horizontal displacement of the projectile:

Therefore, the projectile will land about 17.5\text{ m} from the base of the cliff. 

Example 2: Finding the Maximum Height of a Projectile

As another example, a projectile is launched from the ground with an initial velocity of 25\text{ m/s} at an angle of 50^{\circ} . What is the projectile’s maximum height?

As a projectile travels upward, its vertical velocity becomes slower and slower due to the negative acceleration of gravity. At the maximum height of the trajectory, the projectile’s vertical velocity will momentarily be zero as the projectile stops and turns to move downward. Therefore, in this scenario, our final vertical velocity, v_{fy} , is zero.

We can use the following kinematic equation to solve for the maximum height, d_y :

Solving for d_y :

Before we can use this equation to calculate the height, we will need to use the sine trig identity to find the vertical component of the initial velocity:

Since the initial velocity is in the opposite direction as the acceleration, it’s really important to remember the sign here. If we define moving up as positive, then the initial velocity is positive and the acceleration is negative. Substituting this initial vertical velocity and the given values into the equation above gives:

Therefore, the projectile will reach a maximum height of about 18.7\text{ m} .

For more examples and an explanation of solving these types of projectile motion problems, check out this youtube video from Professor Dave . 

Understanding free fall and projectile motion allows you to solve some of the most complex problems you will encounter in introductory physics. All projectiles are acted on only by gravity, and the vertical and horizontal components of motion are independent of each other. This allows us to apply our kinematic equations to solve for a projectile’s time of flight, velocity, and displacement in each direction.  

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3.5 Free Fall

Learning objectives.

By the end of this section, you will be able to:

• Use the kinematic equations with the variables y and g to analyze free-fall motion.
• Describe how the values of the position, velocity, and acceleration change during a free fall.
• Solve for the position, velocity, and acceleration as functions of time when an object is in a free fall.

An interesting application of Equation 3.4 through Equation 3.14 is called free fall , which describes the motion of an object falling in a gravitational field, such as near the surface of Earth or other celestial objects of planetary size. Let’s assume the body is falling in a straight line perpendicular to the surface, so its motion is one-dimensional. For example, we can estimate the depth of a vertical mine shaft by dropping a rock into it and listening for the rock to hit the bottom. But “falling,” in the context of free fall, does not necessarily imply the body is moving from a greater height to a lesser height. If a ball is thrown upward, the equations of free fall apply equally to its ascent as well as its descent.

The most remarkable and unexpected fact about falling objects is that if air resistance and friction are negligible, then in a given location all objects fall toward the center of Earth with the same constant acceleration , independent of their mass . This experimentally determined fact is unexpected because we are so accustomed to the effects of air resistance and friction that we expect light objects to fall slower than heavy ones. Until Galileo Galilei (1564–1642) proved otherwise, people believed that a heavier object has a greater acceleration in a free fall. We now know this is not the case. In the absence of air resistance, heavy objects arrive at the ground at the same time as lighter objects when dropped from the same height Figure 3.26 .

In the real world, air resistance can cause a lighter object to fall slower than a heavier object of the same size. A tennis ball reaches the ground after a baseball dropped at the same time. (It might be difficult to observe the difference if the height is not large.) Air resistance opposes the motion of an object through the air, and friction between objects—such as between clothes and a laundry chute or between a stone and a pool into which it is dropped—also opposes motion between them.

For the ideal situations of these first few chapters, an object falling without air resistance or friction is defined to be in free fall . The force of gravity causes objects to fall toward the center of Earth. The acceleration of free-falling objects is therefore called acceleration due to gravity . Acceleration due to gravity is constant, which means we can apply the kinematic equations to any falling object where air resistance and friction are negligible. This opens to us a broad class of interesting situations.

Acceleration due to gravity is so important that its magnitude is given its own symbol, g . It is constant at any given location on Earth and has the average value

Although g varies from 9.78 m/s 2 to 9.83 m/s 2 , depending on latitude, altitude, underlying geological formations, and local topography, let’s use an average value of 9.8 m/s 2 rounded to two significant figures in this text unless specified otherwise. Neglecting these effects on the value of g as a result of position on Earth’s surface, as well as effects resulting from Earth’s rotation, we take the direction of acceleration due to gravity to be downward (toward the center of Earth). In fact, its direction defines what we call vertical. Note that whether acceleration a in the kinematic equations has the value + g or − g depends on how we define our coordinate system. If we define the upward direction as positive, then a = − g = −9.8 m/s 2 , a = − g = −9.8 m/s 2 , and if we define the downward direction as positive, then a = g = 9.8 m/s 2 a = g = 9.8 m/s 2 .

One-Dimensional Motion Involving Gravity

The best way to see the basic features of motion involving gravity is to start with the simplest situations and then progress toward more complex ones. So, we start by considering straight up-and-down motion with no air resistance or friction. These assumptions mean the velocity (if there is any) is vertical. If an object is dropped, we know the initial velocity is zero when in free fall. When the object has left contact with whatever held or threw it, the object is in free fall. When the object is thrown, it has the same initial speed in free fall as it did before it was released. When the object comes in contact with the ground or any other object, it is no longer in free fall and its acceleration of g is no longer valid. Under these circumstances, the motion is one-dimensional and has constant acceleration of magnitude g . We represent vertical displacement with the symbol y .

Kinematic Equations for Objects in Free Fall

We assume here that acceleration equals − g (with the positive direction upward).

Problem-Solving Strategy

• Decide on the sign of the acceleration of gravity. In Equation 3.15 through Equation 3.17 , acceleration g is negative, which says the positive direction is upward and the negative direction is downward. In some problems, it may be useful to have acceleration g as positive, indicating the positive direction is downward.
• Draw a sketch of the problem. This helps visualize the physics involved.
• Record the knowns and unknowns from the problem description. This helps devise a strategy for selecting the appropriate equations to solve the problem.
• Decide which of Equation 3.15 through Equation 3.17 are to be used to solve for the unknowns.

Example 3.14

Free fall of a ball.

• Substitute the given values into the equation: y = y 0 + v 0 t − 1 2 g t 2 − 98.0 m = 0 − ( 4.9 m/s ) t − 1 2 ( 9.8 m/s 2 ) t 2 . y = y 0 + v 0 t − 1 2 g t 2 − 98.0 m = 0 − ( 4.9 m/s ) t − 1 2 ( 9.8 m/s 2 ) t 2 . This simplifies to t 2 + t − 20 = 0 . t 2 + t − 20 = 0 . This is a quadratic equation with roots t = −5.0 s and t = 4.0 s t = −5.0 s and t = 4.0 s . The positive root is the one we are interested in, since time t = 0 t = 0 is the time when the ball is released at the top of the building. (The time t = −5.0 s t = −5.0 s represents the fact that a ball thrown upward from the ground would have been in the air for 5.0 s when it passed by the top of the building moving downward at 4.9 m/s.)
• Using Equation 3.15 , we have v = v 0 − g t = −4.9 m/s − ( 9.8 m/s 2 ) ( 4.0 s ) = −44.1 m/s . v = v 0 − g t = −4.9 m/s − ( 9.8 m/s 2 ) ( 4.0 s ) = −44.1 m/s .

Significance

Example 3.15, vertical motion of a baseball.

• Equation 3.16 gives y = y 0 + v 0 t − 1 2 g t 2 y = y 0 + v 0 t − 1 2 g t 2 0 = 0 + v 0 ( 5.0 s ) − 1 2 ( 9.8 m / s 2 ) ( 5.0 s ) 2 , 0 = 0 + v 0 ( 5.0 s ) − 1 2 ( 9.8 m / s 2 ) ( 5.0 s ) 2 , which gives v 0 = 24.5 m/s v 0 = 24.5 m/s .
• At the maximum height, v = 0 v = 0 . With v 0 = 24.5 m/s v 0 = 24.5 m/s , Equation 3.17 gives v 2 = v 0 2 − 2 g ( y − y 0 ) v 2 = v 0 2 − 2 g ( y − y 0 ) 0 = ( 24.5 m/s ) 2 − 2 ( 9.8 m/s 2 ) ( y − 0 ) 0 = ( 24.5 m/s ) 2 − 2 ( 9.8 m/s 2 ) ( y − 0 ) or y = 30.6 m . y = 30.6 m .
• To find the time when v = 0 v = 0 , we use Equation 3.15 : v = v 0 − g t v = v 0 − g t 0 = 24.5 m/s − ( 9.8 m/s 2 ) t . 0 = 24.5 m/s − ( 9.8 m/s 2 ) t . This gives t = 2.5 s t = 2.5 s . Since the ball rises for 2.5 s, the time to fall is 2.5 s.
• The acceleration is 9.8 m/s 2 everywhere, even when the velocity is zero at the top of the path. Although the velocity is zero at the top, it is changing at the rate of 9.8 m/s 2 downward.
• The velocity at t = 5.0 s t = 5.0 s can be determined with Equation 3.15 : v = v 0 − g t = 24.5 m/s − 9.8 m/s 2 ( 5.0 s ) = −24.5 m/s . v = v 0 − g t = 24.5 m/s − 9.8 m/s 2 ( 5.0 s ) = −24.5 m/s .

Check Your Understanding 3.7

A chunk of ice breaks off a glacier and falls 30.0 m before it hits the water. Assuming it falls freely (there is no air resistance), how long does it take to hit the water? Which quantity increases faster, the speed of the ice chunk or its distance traveled?

Example 3.16

Rocket booster.

• From Equation 3.17 , v 2 = v 0 2 − 2 g ( y − y 0 ) v 2 = v 0 2 − 2 g ( y − y 0 ) . With v = 0 and y 0 = 0 v = 0 and y 0 = 0 , we can solve for y : y = v 0 2 2 g = ( 2.0 × 10 2 m / s ) 2 2 ( 9.8 m / s 2 ) = 2040.8 m . y = v 0 2 2 g = ( 2.0 × 10 2 m / s ) 2 2 ( 9.8 m / s 2 ) = 2040.8 m . This solution gives the maximum height of the booster in our coordinate system, which has its origin at the point of release, so the maximum height of the booster is roughly 7.0 km.
• An altitude of 6.0 km corresponds to y = 1.0 × 10 3 m y = 1.0 × 10 3 m in the coordinate system we are using. The other initial conditions are y 0 = 0 , and v 0 = 200.0 m/s y 0 = 0 , and v 0 = 200.0 m/s . We have, from Equation 3.17 , v 2 = ( 200.0 m / s ) 2 − 2 ( 9.8 m / s 2 ) ( 1.0 × 10 3 m ) ⇒ v = ± 142.8 m / s . v 2 = ( 200.0 m / s ) 2 − 2 ( 9.8 m / s 2 ) ( 1.0 × 10 3 m ) ⇒ v = ± 142.8 m / s .

Interactive

Engage the Phet simulation below to learn about graphing polynomials. The shape of the curve changes as the constants are adjusted. View the curves for the individual terms (for example, y = bx ) to see how they add to generate the polynomial curve.

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2.5: Free-Falling Objects

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learning objectives

• Solve basic problems concerning free fall and distinguish it from other kinds of motion

The motion of falling objects is the simplest and most common example of motion with changing velocity. If a coin and a piece of paper are simultaneously dropped side by side, the paper takes much longer to hit the ground. However, if you crumple the paper into a compact ball and drop the items again, it will look like both the coin and the paper hit the floor simultaneously. This is because the amount of force acting on an object is a function of not only its mass, but also area. Free fall is the motion of a body where its weight is the only force acting on an object.

Free Fall : This clip shows an object in free fall.

Galileo also observed this phenomena and realized that it disagreed with the Aristotle principle that heavier items fall more quickly. Galileo then hypothesized that there is an upward force exerted by air in addition to the downward force of gravity. If air resistance and friction are negligible, then in a given location (because gravity changes with location), all objects fall toward the center of Earth with the same constant acceleration , independent of their mass , that constant acceleration is gravity. Air resistance opposes the motion of an object through the air, while friction opposes motion between objects and the medium through which they are traveling. The acceleration of free-falling objects is referred to as the acceleration due to gravity gg. As we said earlier, gravity varies depending on location and altitude on Earth (or any other planet), but the average acceleration due to gravity on Earth is 9.8 \(\mathrm{\frac{m}{s^2}}\). This value is also often expressed as a negative acceleration in mathematical calculations due to the downward direction of gravity.

The best way to see the basic features of motion involving gravity is to start by considering straight up and down motion with no air resistance or friction. This means that if the object is dropped, we know the initial velocity is zero. Once the object is in motion, the object is in free-fall. Under these circumstances, the motion is one-dimensional and has constant acceleration, gg. The kinematic equations for objects experiencing free fall are:

\[\begin{align} \mathrm{v} & \mathrm{=v_0−gt} \\ \mathrm{y} & \mathrm{=y_0+v_0t−\frac{1}{2}gt^2} \\ \mathrm{v^2} & \mathrm{=v^2_0−2g(y−y_0),} \end{align}\]

where \(\mathrm{v=velocity, g=gravity, t=time,}\) and \(\mathrm{y=vertical \; displacement}\).

Video \(\PageIndex{1}\) : Free Fall Motion - Describes how to calculate the time for an object to fall if given the height and the height that an object fell if given the time to fall.

Example \(\PageIndex{1}\):

Some examples of objects that are in free fall include:

• A spacecraft in continuous orbit. The free fall would end once the propulsion devices turned on.
• An stone dropped down an empty well.
• An object, in projectile motion, on its descent.
• The acceleration of free-falling objects is called the acceleration due to gravity, since objects are pulled towards the center of the earth.
• The acceleration due to gravity is constant on the surface of the Earth and has the value of 9.80 \(\mathrm{\frac{m}{s^2}}\) .

The amount by which a speed or velocity changes within a certain period of time (and so a scalar quantity or a vector quantity).

CC LICENSED CONTENT, SHARED PREVIOUSLY

3 Motion Along a Straight Line

3.5 Free Fall

Learning objectives.

By the end of this section, you will be able to:

• Use the kinematic equations with the variables y and g to analyze free-fall motion.
• Describe how the values of the position, velocity, and acceleration change during a free fall.
• Solve for the position, velocity, and acceleration as functions of time when an object is in a free fall.

An interesting application of Figure through Figure is called free fall , which describes the motion of an object falling in a gravitational field, such as near the surface of Earth or other celestial objects of planetary size. Let’s assume the body is falling in a straight line perpendicular to the surface, so its motion is one-dimensional. For example, we can estimate the depth of a vertical mine shaft by dropping a rock into it and listening for the rock to hit the bottom. But “falling,” in the context of free fall, does not necessarily imply the body is moving from a greater height to a lesser height. If a ball is thrown upward, the equations of free fall apply equally to its ascent as well as its descent.

The most remarkable and unexpected fact about falling objects is that if air resistance and friction are negligible, then in a given location all objects fall toward the center of Earth with the same constant acceleration , independent of their mass . This experimentally determined fact is unexpected because we are so accustomed to the effects of air resistance and friction that we expect light objects to fall slower than heavy ones. Until Galileo Galilei (1564–1642) proved otherwise, people believed that a heavier object has a greater acceleration in a free fall. We now know this is not the case. In the absence of air resistance, heavy objects arrive at the ground at the same time as lighter objects when dropped from the same height Figure .

In the real world, air resistance can cause a lighter object to fall slower than a heavier object of the same size. A tennis ball reaches the ground after a baseball dropped at the same time. (It might be difficult to observe the difference if the height is not large.) Air resistance opposes the motion of an object through the air, and friction between objects—such as between clothes and a laundry chute or between a stone and a pool into which it is dropped—also opposes motion between them.

For the ideal situations of these first few chapters, an object falling without air resistance or friction is defined to be in free fall . The force of gravity causes objects to fall toward the center of Earth. The acceleration of free-falling objects is therefore called acceleration due to gravity . Acceleration due to gravity is constant, which means we can apply the kinematic equations to any falling object where air resistance and friction are negligible. This opens to us a broad class of interesting situations.

Acceleration due to gravity is so important that its magnitude is given its own symbol, g . It is constant at any given location on Earth and has the average value

Although g varies from 9.78 m/s 2 to 9.83 m/s 2 , depending on latitude, altitude, underlying geological formations, and local topography, let’s use an average value of 9.8 m/s 2 rounded to two significant figures in this text unless specified otherwise. Neglecting these effects on the value of g as a result of position on Earth’s surface, as well as effects resulting from Earth’s rotation, we take the direction of acceleration due to gravity to be downward (toward the center of Earth). In fact, its direction defines what we call vertical. Note that whether acceleration a in the kinematic equations has the value + g or − g depends on how we define our coordinate system. If we define the upward direction as positive, then [latex]a=\text{−}g=-9.8\,{\text{m/s}}^{2},[/latex] and if we define the downward direction as positive, then [latex]a=g=9.8\,{\text{m/s}}^{2}[/latex].

One-Dimensional Motion Involving Gravity

The best way to see the basic features of motion involving gravity is to start with the simplest situations and then progress toward more complex ones. So, we start by considering straight up-and-down motion with no air resistance or friction. These assumptions mean the velocity (if there is any) is vertical. If an object is dropped, we know the initial velocity is zero when in free fall. When the object has left contact with whatever held or threw it, the object is in free fall. When the object is thrown, it has the same initial speed in free fall as it did before it was released. When the object comes in contact with the ground or any other object, it is no longer in free fall and its acceleration of g is no longer valid. Under these circumstances, the motion is one-dimensional and has constant acceleration of magnitude g . We represent vertical displacement with the symbol y .

Kinematic Equations for Objects in Free Fall

We assume here that acceleration equals − g (with the positive direction upward).

Problem-Solving Strategy: Free Fall

• Decide on the sign of the acceleration of gravity. In Figure through Figure , acceleration g is negative, which says the positive direction is upward and the negative direction is downward. In some problems, it may be useful to have acceleration g as positive, indicating the positive direction is downward.
• Draw a sketch of the problem. This helps visualize the physics involved.
• Record the knowns and unknowns from the problem description. This helps devise a strategy for selecting the appropriate equations to solve the problem.
• Decide which of Figure through Figure are to be used to solve for the unknowns.

Free Fall of a Ball

Figure shows the positions of a ball, at 1-s intervals, with an initial velocity of 4.9 m/s downward, that is thrown from the top of a 98-m-high building. (a) How much time elapses before the ball reaches the ground? (b) What is the velocity when it arrives at the ground?

Choose the origin at the top of the building with the positive direction upward and the negative direction downward. To find the time when the position is −98 m, we use Figure , with [latex]{y}_{0}=0,{v}_{0}=-4.9\,\text{m/s,}\,\text{and}\,g=9.8\,{\text{m/s}}^{2}[/latex].

Substitute the given values into the equation:

[latex]\begin{array}{cc} y={y}_{0}+{v}_{0}t-\frac{1}{2}g{t}^{2}\hfill \\ -98.0\,\text{m}=0-(4.9\,\text{m/s})t-\frac{1}{2}(9.8\,{\text{m/s}}^{2}){t}^{2}.\hfill \end{array}[/latex]

This simplifies to

[latex]{t}^{2}+t-20=0.[/latex]

This is a quadratic equation with roots [latex]t=-5.0\mathrm{s}\,\text{and}\,t=4.0\mathrm{s}[/latex]. The positive root is the one we are interested in, since time [latex]t=0[/latex] is the time when the ball is released at the top of the building. (The time [latex]t=-5.0\mathrm{s}[/latex] represents the fact that a ball thrown upward from the ground would have been in the air for 5.0 s when it passed by the top of the building moving downward at 4.9 m/s.)

Using (Figure), we have

[latex]v={v}_{0}-gt=-4.9\,\text{m/s}-(9.8{\text{m/s}}^{2})(4.0\,\text{s})=-44.1\,\text{m/s}\text{.}[/latex]

Significance

For situations when two roots are obtained from a quadratic equation in the time variable, we must look at the physical significance of both roots to determine which is correct. Since [latex]t=0[/latex] corresponds to the time when the ball was released, the negative root would correspond to a time before the ball was released, which is not physically meaningful. When the ball hits the ground, its velocity is not immediately zero, but as soon as the ball interacts with the ground, its acceleration is not g and it accelerates with a different value over a short time to zero velocity. This problem shows how important it is to establish the correct coordinate system and to keep the signs of g in the kinematic equations consistent.

Vertical Motion of a Baseball

A batter hits a baseball straight upward at home plate and the ball is caught 5.0 s after it is struck Figure . (a) What is the initial velocity of the ball? (b) What is the maximum height the ball reaches? (c) How long does it take to reach the maximum height? (d) What is the acceleration at the top of its path? (e) What is the velocity of the ball when it is caught? Assume the ball is hit and caught at the same location.

Choose a coordinate system with a positive y -axis that is straight up and with an origin that is at the spot where the ball is hit and caught.

Figure gives

which gives [latex]{v}_{0}=24.5\,\text{m/sec}[/latex].

At the maximum height, [latex]v=0[/latex]. With [latex]{v}_{0}=24.5\,\text{m/s}[/latex], Figure gives

To find the time when [latex]v=0[/latex], we use Figure :

This gives [latex]t=2.5\,\text{s}[/latex]. Since the ball rises for 2.5 s, the time to fall is 2.5 s.

The acceleration is 9.8 m/s2 everywhere, even when the velocity is zero at the top of the path. Although the velocity is zero at the top, it is changing at the rate of 9.8 m/s2 downward.

The velocity at [latex]t=5.0\mathrm{s}[/latex] can be determined with (Figure): [latex]\begin{array}{cc}\hfill v& ={v}_{0}-gt\hfill \\ & =24.5\,\text{m/s}-9.8{\text{m/s}}^{2}(5.0\,\text{s})\hfill \\ & =-24.5\,\text{m/s}.\hfill \end{array}[/latex]

The ball returns with the speed it had when it left. This is a general property of free fall for any initial velocity. We used a single equation to go from throw to catch, and did not have to break the motion into two segments, upward and downward. We are used to thinking of the effect of gravity is to create free fall downward toward Earth. It is important to understand, as illustrated in this example, that objects moving upward away from Earth are also in a state of free fall.

Check Your Understanding

A chunk of ice breaks off a glacier and falls 30.0 m before it hits the water. Assuming it falls freely (there is no air resistance), how long does it take to hit the water? Which quantity increases faster, the speed of the ice chunk or its distance traveled?

It takes 2.47 s to hit the water. The quantity distance traveled increases faster.

Rocket Booster

A small rocket with a booster blasts off and heads straight upward. When at a height of [latex]5.0\,\text{km}[/latex] and velocity of 200.0 m/s, it releases its booster. (a) What is the maximum height the booster attains? (b) What is the velocity of the booster at a height of 6.0 km? Neglect air resistance.

We need to select the coordinate system for the acceleration of gravity, which we take as negative downward. We are given the initial velocity of the booster and its height. We consider the point of release as the origin. We know the velocity is zero at the maximum position within the acceleration interval; thus, the velocity of the booster is zero at its maximum height, so we can use this information as well. From these observations, we use Figure , which gives us the maximum height of the booster. We also use Figure to give the velocity at 6.0 km. The initial velocity of the booster is 200.0 m/s.

From Figure , [latex]{v}^{2}={v}_{0}^{2}-2g(y-{y}_{0})[/latex]. With [latex]v=0\,\text{and}\,{y}_{0}=0[/latex], we can solve for y:

[latex]y=\frac{{v}_{0}^{2}}{-2g}=\frac{(2.0\times {10}^{2}\text{m}\text{/}{\text{s})}^{2}}{-2(9.8\,\text{m}\text{/}{\text{s}}^{2})}=2040.8\,\text{m}\text{.}[/latex]

This solution gives the maximum height of the booster in our coordinate system, which has its origin at the point of release, so the maximum height of the booster is roughly 7.0 km.

An altitude of 6.0 km corresponds to[latex]y=1.0\times {10}^{3}\,\text{m}[/latex] in the coordinate system we are using. The other initial conditions are[latex]{y}_{0}=0,\,\text{and}\,{v}_{0}=200.0\,\text{m/s}[/latex].

We have, from Figure ,

[latex]{v}^{2}={(200.0\,\text{m}\text{/}\text{s})}^{2}-2(9.8\,\text{m}\text{/}{\text{s}}^{2})(1.0\times {10}^{3}\,\text{m})\Rightarrow v=\pm142.8\,\text{m}\text{/}\text{s}.[/latex]

We have both a positive and negative solution in (b). Since our coordinate system has the positive direction upward, the +142.8 m/s corresponds to a positive upward velocity at 6000 m during the upward leg of the trajectory of the booster. The value v = −142.8 m/s corresponds to the velocity at 6000 m on the downward leg. This example is also important in that an object is given an initial velocity at the origin of our coordinate system, but the origin is at an altitude above the surface of Earth, which must be taken into account when forming the solution.

Visit this site to learn about graphing polynomials. The shape of the curve changes as the constants are adjusted. View the curves for the individual terms (for example, y = bx ) to see how they add to generate the polynomial curve.

• An object in free fall experiences constant acceleration if air resistance is negligible.
• On Earth, all free-falling objects have an acceleration g due to gravity, which averages [latex]g=9.81\,{\text{m/s}}^{2}[/latex].
• For objects in free fall, the upward direction is normally taken as positive for displacement, velocity, and acceleration.

Conceptual Questions

What is the acceleration of a rock thrown straight upward on the way up? At the top of its flight? On the way down? Assume there is no air resistance.

An object that is thrown straight up falls back to Earth. This is one-dimensional motion. (a) When is its velocity zero? (b) Does its velocity change direction? (c) Does the acceleration have the same sign on the way up as on the way down?

a. at the top of its trajectory; b. yes, at the top of its trajectory; c. yes

Suppose you throw a rock nearly straight up at a coconut in a palm tree and the rock just misses the coconut on the way up but hits the coconut on the way down. Neglecting air resistance and the slight horizontal variation in motion to account for the hit and miss of the coconut, how does the speed of the rock when it hits the coconut on the way down compare with what it would have been if it had hit the coconut on the way up? Is it more likely to dislodge the coconut on the way up or down? Explain.

The severity of a fall depends on your speed when you strike the ground. All factors but the acceleration from gravity being the same, how many times higher could a safe fall on the Moon than on Earth (gravitational acceleration on the Moon is about one-sixth that of the Earth)?

Earth [latex]v={v}_{0}-gt=\text{−}gt[/latex]; Moon [latex]{v}^{\prime }=\frac{g}{6}{t}^{\prime }v={v}^{\prime }\enspace -gt=-\frac{g}{6}{t}^{\prime }\enspace{t}^{\prime }=6t[/latex]; Earth [latex]y=-\frac{1}{2}g{t}^{2}[/latex] Moon [latex]{y}^{\prime }=-\frac{1}{2}\,\frac{g}{6}{(6t)}^{2}=-\frac{1}{2}g6{t}^{2}=-6(\frac{1}{2}g{t}^{2})=-6y[/latex]

How many times higher could an astronaut jump on the Moon than on Earth if her takeoff speed is the same in both locations (gravitational acceleration on the Moon is about on-sixth of that on Earth)?

Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, and (d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. Take the point of release to be [latex]{y}_{0}=0[/latex].

Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, (d) 2.00 s, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.

a. [latex]\begin{array}{cc} y=-8.23\,\text{m}\hfill \\ {v}_{1}=\text{−}18.9\,\text{m/s}\hfill \end{array}[/latex];

b. [latex]\begin{array}{cc} y=-18.9\,\text{m}\hfill \\ {v}_{2}=23.8\,\text{m/s}\hfill \end{array}[/latex];

c. [latex]\begin{array}{cc} y=-32.0\,\text{m}\hfill \\ {v}_{3}=\text{−}28.7\,\text{m/s}\hfill \end{array}[/latex];

d. [latex]\begin{array}{cc} y=-47.6\,\text{m}\hfill \\ {v}_{4}=\text{−}33.6\,\text{m/s}\hfill \end{array}[/latex];

e. [latex]\begin{array}{cc} y=-65.6\,\text{m}\hfill \\ {v}_{5}=\text{−}38.5\,\text{m/s}\hfill \end{array}[/latex]

A basketball referee tosses the ball straight up for the starting tip-off. At what velocity must a basketball player leave the ground to rise 1.25 m above the floor in an attempt to get the ball?

A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of 1.40 m/s and observes that it takes 1.8 s to reach the water. (a) List the knowns in this problem. (b) How high above the water was the preserver released? Note that the downdraft of the helicopter reduces the effects of air resistance on the falling life preserver, so that an acceleration equal to that of gravity is reasonable.

a. Knowns: [latex]a=\text{−}9.8\,{\text{m/s}}^{2}\enspace{v}_{0}=-1.4\,\text{m/s}t=1.8\,\text{s}\enspace{y}_{0}=0\,\text{m}[/latex];

b. [latex]y={y}_{0}+{v}_{0}t-\frac{1}{2}g{t}^{2}y={v}_{0}t-\frac{1}{2}gt=-1.4\,\text{m}\text{/}\text{s}(1.8\,\text{sec})-\frac{1}{2}(9.8){(1.8\,\text{s})}^{2}=-18.4\,\text{m}[/latex] and the origin is at the rescuers, who are 18.4 m above the water.

Unreasonable results A dolphin in an aquatic show jumps straight up out of the water at a velocity of 15.0 m/s. (a) List the knowns in this problem. (b) How high does his body rise above the water? To solve this part, first note that the final velocity is now a known, and identify its value. Then, identify the unknown and discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking units, and discuss whether the answer is reasonable. (c) How long a time is the dolphin in the air? Neglect any effects resulting from his size or orientation.

A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet first into a pool. She starts with a velocity of 4.00 m/s and her takeoff point is 1.80 m above the pool. (a) What is her highest point above the board? (b) How long a time are her feet in the air? (c) What is her velocity when her feet hit the water?

a. [latex]{v}^{2}={v}_{0}^{2}-2g(y-{y}_{0})\enspace{y}_{0}=0v=0y=\frac{{v}_{0}^{2}}{2g}=\frac{{(4.0\,\text{m}\text{/}\text{s})}^{2}}{2(9.80)}=0.82\,\text{m}[/latex]; b. to the apex [latex]v=0.41\,\text{s}[/latex] times 2 to the board = 0.82 s from the board to the water [latex]y={y}_{0}+{v}_{0}t-\frac{1}{2}g{t}^{2}y=-1.80\,\text{m}\enspace{y}_{0}=0\enspace{v}_{0}=4.0\,\text{m}\text{/}\text{s}[/latex] [latex]-1.8=4.0t-4.9{t}^{2}\enspace 4.9{t}^{2}-4.0t-1.80=0[/latex], solution to quadratic equation gives 1.13 s; c. [latex]\begin{array}{c}{v}^{2}={v}_{0}^{2}-2g(y-{y}_{0})\enspace{y}_{0}=0\enspace\enspace{v}_{0}=4.0\,\text{m}\text{/}\text{s}y=-1.80\,\text{m}\hfill \\ v=7.16\,\text{m}\text{/}\text{s}\hfill \end{array}[/latex]

(a) Calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.00 m/s. (b) How long a time would it take to reach the ground if it is thrown straight down with the same speed?

A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. How long a time does he have to get out of the way if the shot was released at a height of 2.20 m and he is 1.80 m tall?

Time to the apex: [latex]t=1.12\,\text{s}[/latex] times 2 equals 2.24 s to a height of 2.20 m. To 1.80 m in height is an additional 0.40 m. [latex]\begin{array}{cc} y={y}_{0}+{v}_{0}t-\frac{1}{2}g{t}^{2}y=-0.40\,\text{m}\enspace{y}_{0}=0\enspace{v}_{0}=-11.0\,\text{m}\text{/}\text{s}\hfill \\ y={y}_{0}+{v}_{0}t-\frac{1}{2}g{t}^{2}y=-0.40\,\text{m}\enspace{y}_{0}=0\enspace{v}_{0}=-11.0\,\text{m}\text{/}\text{s}\hfill \\ -0.40=-11.0t-4.9{t}^{2}\enspace\text{or}\enspace 4.9{t}^{2}+11.0t-0.40=0\hfill \end{array}[/latex].

Take the positive root, so the time to go the additional 0.4 m is 0.04 s. Total time is [latex]2.24\,\text{s}\,+0.04\,\text{s}\,=2.28\,\text{s}[/latex].

You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.0 m. How much additional time elapses before the ball passes the tree branch on the way back down?

A kangaroo can jump over an object 2.50 m high. (a) Considering just its vertical motion, calculate its vertical speed when it leaves the ground. (b) How long a time is it in the air?

a. [latex]\begin{array}{cc} {v}^{2}={v}_{0}^{2}-2g(y-{y}_{0})\enspace{y}_{0}=0v=0y=2.50\,\text{m}\hfill \\ {v}_{0}^{2}=2gy\Rightarrow {v}_{0}=\sqrt{2(9.80)(2.50)}=7.0\,\text{m}\text{/}\text{s}\hfill \end{array}[/latex]; b. [latex]t=0.72\,\text{s}[/latex] times 2 gives 1.44 s in the air

Standing at the base of one of the cliffs of Mt. Arapiles in Victoria, Australia, a hiker hears a rock break loose from a height of 105.0 m. He can’t see the rock right away, but then does, 1.50 s later. (a) How far above the hiker is the rock when he can see it? (b) How much time does he have to move before the rock hits his head?

There is a 250-m-high cliff at Half Dome in Yosemite National Park in California. Suppose a boulder breaks loose from the top of this cliff. (a) How fast will it be going when it strikes the ground? (b) Assuming a reaction time of 0.300 s, how long a time will a tourist at the bottom have to get out of the way after hearing the sound of the rock breaking loose (neglecting the height of the tourist, which would become negligible anyway if hit)? The speed of sound is 335.0 m/s on this day.

a. [latex]v=70.0\,\text{m}\text{/}\text{s}[/latex]; b. time heard after rock begins to fall: 0.75 s, time to reach the ground: 6.09 s

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Free Fall Motion: Tutorials with Examples and Solutions

Problems on free fall motion are presented along with detailed solutions.

From rest, a car accelerated at 8 m/s 2 for 10 seconds. a) What is the position of the car at the end of the 10 seconds? b) What is the velocity of the car at the end of the 10 seconds? Solution to Problem 1

With an initial velocity of 20 km/h, a car accelerated at 8 m/s 2 for 10 seconds. a) What is the position of the car at the end of the 10 seconds? b) What is the velocity of the car at the end of the 10 seconds? Solution to Problem 2

A car accelerates uniformly from 0 to 72 km/h in 11.5 seconds. a) What is the acceleration of the car in m/s 2 ? b) What is the position of the car by the time it reaches the velocity of 72 km/h? Solution to Problem 3

An object is thrown straight down from the top of a building at a speed of 20 m/s. It hits the ground with a speed of 40 m/s. a) How high is the building? b) How long was the object in the air? Solution to Problem 4

A train brakes from 40 m/s to a stop over a distance of 100 m. a) What is the acceleration of the train? b) How much time does it take the train to stop? Solution to Problem 5

A boy on a bicycle increases his velocity from 5 m/s to 20 m/s in 10 seconds. a) What is the acceleration of the bicycle? b) What distance was covered by the bicycle during the 10 seconds? Solution to Problem 6

a) How long does it take an airplane to take off if it needs to reach a speed on the ground of 350 km/h over a distance of 600 meters (assume the plane starts from rest)? b) What is the acceleration of the airplane over the 600 meters? Solution to Problem 7

Starting from a distance of 20 meters to the left of the origin and at a velocity of 10 m/s, an object accelerates to the right of the origin for 5 seconds at 4 m/s 2 . What is the position of the object at the end of the 5 seconds of acceleration? Solution to Problem 8

What is the smallest distance, in meters, needed for an airplane touching the runway with a velocity of 360 km/h and an acceleration of -10 m/s 2 to come to rest? Solution to Problem 9

Problem 10:

To approximate the height of a water well, Martha and John drop a heavy rock into the well. 8 seconds after the rock is dropped, they hear a splash caused by the impact of the rock on the water. What is the height of the well. (Speed of sound in air is 340 m/s). Solution to Problem 10

Problem 11:

A rock is thrown straight up and reaches a height of 10 m. a) How long was the rock in the air? b) What is the initial velocity of the rock? Solution to Problem 11

Problem 12:

A car accelerates from rest at 1.0 m/s 2 for 20.0 seconds along a straight road . It then moves at a constant speed for half an hour. It then decelerates uniformly to a stop in 30.0 s. Find the total distance covered by the car. Solution to Problem 12

More References and links

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Application: Falling Objects (1D)

Application: projectile motion (2d), forces & newton's laws of motion, potential energy, thermal energy.

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Free Fall (Physics): Definition, Formula, Problems & Solutions (w/ Examples)

​ Free fall ​ refers to situations in physics where the only force acting on an object is gravity.

The simplest examples occur when objects fall from a given height above the surface of the Earth straight downward – a one-dimensional problem. If the object is tossed upward or forcefully thrown straight downward, the example is still one-dimensional, but with a twist.

Projectile motion is a classic category of free-fall problems. In reality, of course, these events unfold in the three-dimensional world, but for introductory physics purposes, they are treated on paper (or on your screen) as two-dimensional: ​ x ​ for right and left (with right being positive), and ​ y ​ for up and down (with up being positive).

Free-fall examples therefore often have negative values for y-displacement.

It is perhaps counterintuitive that some free-fall problems qualify as such.

Keep in mind that the only criterion is that the only force acting on the object is gravity (usually Earth's gravity). Even if an object is launched into the sky with colossal initial force, at the moment the object is released and thereafter, the only force acting on it is gravity and it is now a projectile.

• Often, high-school and many college physics problems neglect air resistance, though this always has at least a slight effect in reality; the exception is an event that unfolds in a vacuum. This is discussed in detail later.

The Unique Contribution of Gravity

A unique an interesting property of the acceleration due to gravity is that it is the same for all masses.

This was far from self-evident until the days of Galileo Galilei (1564-1642). That's because in reality gravity is not the only force acting as an object falls, and the effects of air resistance tend to cause lighter objects to accelerate more slowly – something we've all noticed when comparing the fall rate of a rock and a feather.

Galileo conducted ingenious experiments at the "leaning" Tower of Pisa, proving by dropping masses of different weights from the high top of the tower that gravitational acceleration is independent of mass.

Solving Free-Fall Problems

Usually, you are looking to determine initial velocity (v 0y ), final velocity (v y ) or how far something has fallen (y − y 0 ). Although Earth's gravitational acceleration is a constant 9.8 m/s 2 , elsewhere (such as on the moon) the constant acceleration experienced by an object in free fall has a different value.

For free fall in one dimension (for example, an apple falling straight down from a tree), use the kinematic equations in the ​ Kinematic Equations for Free-Falling Objects ​ section. For a projectile-motion problem in two dimensions, use the kinematic equations in the section ​ Projectile Motion and Coordinate Systems ​.

• You can also use the conservation of energy principle, which states that ​ the loss of potential energy (PE) ​ during the fall ​ equals the gain in kinetic energy (KE): ​ –mg(y − y 0 ) = (1/2)mv y 2 . 

Kinematic Equations for Free-Falling Objects

All of the foregoing can be reduced for present purposes to the following three equations. These are tailored for free fall, so that the "y" subscripts can be omitted. Assume that acceleration, per physics convention, equals −g (with the positive direction therefore upward).

• Note that v 0 and y 0  are initial values in any problem, not variables.

​ Example 1: ​ A strange birdlike animal is hovering in the air 10 m directly over your head, daring you to hit it with the rotten tomato you're holding. With what minimum initial velocity v 0 would you have to throw the tomato straight up in order to ensure that it reaches its squawking target?

What's happening physically is that the ball is coming to a stop owing to the force of gravity just as it reaches the required height, so here, v y = v = 0.

First, list your known quantities: ​ v = ​ 0​ , g = ​ –9.8 m/s2​ , y − y 0 = ​ 10 m

Thus you can use the third of the equations above to solve:

This is about 31 miles an hour.

Projectile Motion and Coordinate Systems

Projectile motion involves the motion of an object in (usually) two dimensions under the force of gravity. The behavior of the object in the x-direction and in the y-direction can be described separately in assembling the greater picture of the particle's motion. This means that "g" appears in most of the equations required to solve all projectile-motion problems, not merely those involving free fall.

The kinematic equations needed to solve basic projectile motion problems, which omit air resistance:

​ Example 2: ​ A daredevil decides to try to drive his "rocket car" across the gap between adjacent building rooftops. These are separated by 100 horizontal meters, and the roof of the "take-off" building is 30 m higher than the second (this almost 100 feet, or perhaps 8 to 10 "floors," i.e., levels).

Neglecting air resistance, how fast will he need to be going as he leaves the first rooftop to assure just reaching the second rooftop? Assume his vertical velocity is zero at the instant the car takes off.

Again, list your known quantities: (x – x 0 ) = 100m, (y – y 0 ) = –30m, v 0y = 0, g = –9.8 m/s 2 .

Here, you take advantage of the fact that horizontal motion and vertical motion can be assessed independently. How long the car will take to free-fall (for purposes of y-motion) 30 m? The answer is given by y – y 0 = v 0y t − (1/2)gt 2.

Filling in the known quantities and solving for t:

Now plug this value into x = x 0 + v 0x t :

​ v 0x = 40.4 m/s (about 90 miles per hour). ​

This is perhaps possible, depending on the size of the roof, but all in all not a good idea outside of action-hero movies.

Hitting it out of the Park... Far Out

Air resistance plays a major, under-appreciated role in everyday events even when free fall is only part of the physical story. In 2018, a professional baseball player named Giancarlo Stanton hit a pitched ball hard enough to blast it away from home plate at a record 121.7 miles per hour.

The equation for the maximum horizontal distance a launched projectile can attain, or ​ range equation ​ (see Resources), is:

Based on this, if Stanton had hit the ball at the theoretical ideal angle of 45 degrees (where sin 2θ is at its maximum value of 1), the ball would have traveled 978 feet! In reality, home runs almost never reach even 500 feet. Part if this is because a launch angle of 45 degrees for a batter is not ideal, as the pitch is coming in almost horizontally. But much of the difference is owed to the velocity-dampening effects of air resistance.

Air Resistance: Anything But "Negligible"

Free-fall physics problems aimed at less advanced students assume the absence of air resistance because this factor would introduce another force that can slow or decelerate objects and would need to be mathematically accounted for. This is a task best reserved for advanced courses, but it bears discussion here nonetheless.

In the real world, the Earth's atmosphere provides some resistance to an object in free fall. Particles in the air collide with the falling object, which results in transforming some of its kinetic energy into thermal energy. Since energy is conserved in general, this results in "less motion" or a more slowly increasing downward velocity.

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About the Author

Kevin Beck holds a bachelor's degree in physics with minors in math and chemistry from the University of Vermont. Formerly with ScienceBlogs.com and the editor of "Run Strong," he has written for Runner's World, Men's Fitness, Competitor, and a variety of other publications. More about Kevin and links to his professional work can be found at www.kemibe.com.

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• Physics Formulas

Free Fall Formula

Freefall as the term says, is a body falling freely because of the gravitational pull of our earth.

Imagine a body with velocity (v) is falling freely from a height (h) for time (t) seconds because of gravity (g).

Free Fall Formulas  are articulated as follows:

h = (1/2) gt 2

Freefall Related Solved Examples

Underneath are given questions on free fall which may be useful for you.

Problem 1:  Calculate  the  body  height  if  it  has  a  mass  of  2  kg  and  after  7 seconds  it  reaches  the  ground?

Given: Height h =? Time t = 7s We all are acquainted with the fact that free fall is independent of mass.

Hence, it is given as

h = 0.5 × 9.8 × (7) 2

h = 240.1 m

Problem  2: The cotton falls after 3 s and iron falls after 5 s. Which is moving with higher velocity? Answer:

The Velocity in free fall is autonomous of mass.

V (Velocity of iron) = gt = 9.8 m/s 2  × 5s = 49 m/s

V (Velocity of cotton) = gt = 9.8 m/s 2  × 3s = 29.4 m/s.

The Velocity of iron is more than cotton.

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1.1: Free Fall

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• Page ID 91046

• Russell Herman
• University of North Carolina Wilmington

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In this chapter we will study some common differential equations that appear in physics. We will begin with the simplest types of equations and standard techniques for solving them We will end this part of the discussion by returning to the problem of free fall with air resistance. We will then turn to the study of oscillations, which are modeled by second order differential equations.

Free fall example.

Let us begin with a simple example from introductory physics. Recall that free fall is the vertical motion of an object solely under the force of gravity. It has been experimentally determined that an object near the surface of the Earth falls at a constant acceleration in the absence of other forces, such as air resistance. This constant acceleration is denoted by \(-g\) , where \(g\) is called the acceleration due to gravity. The negative sign is an indication that we have chosen a coordinate system in which up is positive.

We are interested in determining the position, \(y(t)\) , of the falling body as a function of time. From the definition of free fall, we have

\[\ddot{y}(t)=-g \label{1.1}. \]

Note that we will occasionally use a dot to indicate time differentiation.

Differentiation with respect to time is often denoted by dots instead of primes.

This notation is standard in physics and we will begin to introduce you to this notation, though at times we might use the more familiar prime notation to indicate spatial differentiation, or general differentiation.

In Equation \(\PageIndex{1}\) we know \(g\) . It is a constant. Near the Earth’s surface it is about \(9.81 \mathrm{~m} / \mathrm{s}^{2}\) or \(32.2 \mathrm{ft} / \mathrm{s}^{2}\) . What we do not know is \(y(t)\) . This is our first differential equation. In fact it is natural to see differential equations appear in physics often through Newton’s Second Law, \(F=m a\) , as it plays an important role in classical physics. We will return to this point later.

So, how does one solve the differential equation in \(\PageIndex{1}\)? We do so by using what we know about calculus. It might be easier to see when we put in a particular number instead of \(g\) . You might still be getting used to the fact that some letters are used to represent constants. We will come back to the more general form after we see how to solve the differential equation.

\[\ddot{y}(t)=5. \nonumber \]

Recalling that the second derivative is just the derivative of a derivative, we can rewrite this equation as

\[\dfrac{d}{d t}\left(\dfrac{d y}{d t}\right)=5 \nonumber \]

This tells us that the derivative of \(d y / d t\) is 5 . Can you think of a function whose derivative is 5 ? (Do not forget that the independent variable is \(t\) .) Yes, the derivative of \(5 t\) with respect to \(t\) is 5 . Is this the only function whose derivative is 5 ? No! You can also differentiate \(5 t+1,5 t+\pi, 5 t-6\) , etc. In general, the derivative of \(5 t+C\) is 5, where \(C\) is an arbitrary integration constant.

So, Equation \(\PageIndex{2}\) can be reduced to

\[\dfrac{d y}{d t}=5 t+C \nonumber \]

Now we ask if you know a function whose derivative is \(5 t+C\) . Well, you might be able to do this one in your head, but we just need to recall the Fundamental Theorem of Calculus, which relates integrals and derivatives. Thus, we have

\[y(t)=\dfrac{5}{2} t^{2}+C t+D \nonumber \]

where \(D\) is a second integration constant.

Equation \(\PageIndex{5}\) gives the solution to the original differential equation. That means that when the solution is placed into the differential equation, both sides of the differential equation give the same expression. You can always check your answer to a differential equation by showing that your solution satisfies the equation. In this case we have

\(\ddot{y}(t)=\dfrac{d^{2}}{d t^{2}}\left(\dfrac{5}{2} t^{2}+C t+D\right)=\dfrac{d}{d t}(5 t+C)=5\)

Therefore, Equation \(\PageIndex{5}\) gives the general solution of the differential equation.

We also see that there are two arbitrary constants, \(C\) and \(D .\) Picking any values for these gives a whole family of solutions. As we will see, the equation \(\ddot{y}(t)=5\) is a linear second order ordinary differential equation. The general solution of such an equation always has two arbitrary constants.

Let’s return to the free fall problem. We solve it the same way. The only difference is that we can replace the constant 5 with the constant \(-g .\) So, we find that

\[\dfrac{d y}{d t}=-g t+C \nonumber \]

\[y(t)=-\dfrac{1}{2} g t^{2}+C t+D \nonumber \]

Once you get down the process, it only takes a line or two to solve.

There seems to be a problem. Imagine dropping a ball that then undergoes free fall. We just determined that there are an infinite number of solutions for the position of the ball at any time! Well, that is not possible. Experience tells us that if you drop a ball you expect it to behave the same way every time. Or does it? Actually, you could drop the ball from anywhere. You could also toss it up or throw it down. So, there are many ways you can release the ball before it is in free fall producing many different paths, \(y(t)\) . That is where the constants come in. They have physical meanings.

If you set \(t=0\) in the equation, then you have that \(y(0)=D .\) Thus, \(D\) gives the initial position of the ball. Typically, we denote initial values with a subscript. So, we will write \(y(0)=y_{0}\) . Thus, \(D=y_{0}\) .

That leaves us to determine \(C\) . It appears at first in Equation \(\PageIndex{6}\). Recall that \(\dfrac{d y}{d t}\) , the derivative of the position, is the vertical velocity, \(v(t)\) . It is positive when the ball moves upward. We will denote the initial velocity \(v(0)=v_{0} .\) Inserting \(t=0\) in Equation \(\PageIndex{6}\), we find that \(\dot{y}(0)=C\) . This implies that \(C=v(0)=v_{0}\) .

Putting this all together, we have the physical form of the solution for free fall as

\[y(t)=-\dfrac{1}{2} g t^{2}+v_{0} t+y_{0} \nonumber \]

Doesn’t this equation look familiar? Now we see that the infinite family of solutions consists of free fall resulting from initially dropping a ball at position \(y_{0}\) with initial velocity \(v_{0}\) . The conditions \(y(0)=y_{0}\) and \(\dot{y}(0)=v_{0}\) are called the initial conditions. A solution of a differential equation satisfying a set of initial conditions is often called a particular solution. Specifying the initial conditions results in a unique solution.

So, we have solved the free fall equation. Along the way we have begun to see some of the features that will appear in the solutions of other problems that are modeled with differential equation. Throughout the book we will see several applications of differential equations. We will extend our analysis to higher dimensions, in which we case will be faced with socalled partial differential equations, which involve the partial derivatives of functions of more that one variable.

But are we done with free fall? Not at all! We can relax some of the conditions that we have imposed. We can add air resistance. We will visit this problem later in this chapter after introducing some more techniques. We can also provide a horizontal component of motion, leading to projectile motion.

Finally, we should also note that free fall at constant \(g\) only takes place near the surface of the Earth. What if a tile falls off the shuttle far from the surface of the Earth? It will also fall towards the Earth. Actually, the tile also has a velocity component in the direction of the motion of the shuttle. So, it would not necessarily take radial path downwards. For now, let’s ignore that component. To look at this problem in more detail, we need to go to the origins of the acceleration due to gravity. This comes out of Newton’s Law of Gravitation. Consider a mass \(m\) at some distance \(h(t)\) from the surface of the (spherical) Earth. Letting \(M\) and \(R\) be the Earth’s mass and radius, respectively, Newton’s Law of Gravitation states that

\[ \begin{aligned} m a &=F \\ m \dfrac{d^{2} h(t)}{d t^{2}} &=-G \dfrac{m M}{(R+h(t))^{2}} \end{aligned} \nonumber \]

Here \(G=6.6730 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}\) is the Universal Gravitational Constant, \(M=5.9736 \times 10^{24} \mathrm{~kg}\) and \(R=6371 \mathrm{~km}\) are the Earth’s mass and mean radius, respectively. For \(h<<R, G M / R^{2} \approx g\) .

Thus, we arrive at a differential equation

\[\dfrac{d^{2} h(t)}{d t^{2}}=-\dfrac{G M}{(R+h(t))^{2}} . \nonumber \]

This equation is not as easy to solve. We will leave it as a homework exercise for the reader.

Gurumuda Networks

Free fall motion – problems and solutions

1. A stone free fall from the height of 45 meters. If the acceleration due to gravity is 10 ms -2 , what is the speed of the stone when hits the ground?

Height (h) = 45 meters

Acceleration due to gravity (g) = 10 m/s 2

Wanted : The final velocity of the stone when it hits the ground (v t )

The equation of free fall motion :

v t 2 = 2 g h

The final velocity of the stone :

v t 2 = 2 (10)(45) = 900

v t = √900 = 30 m/s 2

2. An object free fall from a height without the initial velocity. The object hits the ground 2 seconds later. Acceleration due to gravity is 10 ms -2 . Determine height

Time interval (t) = 2 seconds

Wanted : Height (h)

The equation of the free fall motion :

h = ½ g t 2

h = ½ (10)(2) 2 = (5)(4) = 20 meters

3.A 2-kg object free fall from a height of 20 meters above the ground. What is the time interval the object in air ? Acceleration due to gravity is 10 ms -2

Height (h) = 20 meters

Wanted : Time interval (t)

Time interval :

20 = ½ (10)(t 2 )

20 = (5)(t 2 )

t = 2 seconds

4. Two objects, object 1 and object 2, are free fall from a height of h 1 and h 2 at the same time. If h 1 : h 2 = 2: 1, what is the ratio of the time interval of the object 1 to the object 2.

The height of the object 1 (h 1 ) = 2

The height of the object 2 (h 2 ) = 1

Acceleration due to gravity = g

Wanted : t 1 : t 2

h 1 = 1/2 g t 1 2

2 = 1/2 g t 1 2

(2)(2) = g t 1 2

4 = g t 1 2

4/g = t 1 2

h 2 = 1/2 g t 2 2

1 = 1/2 g t 2 2

(2)(1) = g t 2 2

2 = g t 2 2

2/g = t 2 2

The ratio of the time interval :

√4/g : √2/g

(√4/g) 2 : (√2/g) 2

5. An object dropped from a height of h above the ground. The final velocity when object hits the ground is 10 m/s. What is the time interval to reach ½ h above the ground. Acceleration due to gravity is 10 m/s 2 .

The final velocity (v t ) = 10 m/s

Wanted : The time interval to reach 1/2 h above the ground

The height of h :

10 2 = 2 (10) h

h = 100 / 20

h = 5 meters

The height of 1/2 h = 1/2 (5 meters) = 2.5 meters. The time interval needed to reach 2.5 meters above the ground :

h = 1/2 g t 2

2.5 = 1/2 (10) t 2

2.5 = 5 t 2

t 2 = 2.5 / 5 = 0.5 = (0.25)(2)

t = √(0.25)(2) = 0.5√2 = 1/2 √2 seconds

T he free fall motion of coconut ( figure 1 ) and the motion of a ball thrown vertically up ward to the highest poin t by a student (figur e 2). Determine the kind of both motion s.

Figure 1 = free fall motion = Acceleration

Figure 2 = vertical motion = Deceleration

The correct answer is A.

7. A stone free fall from a building. The time interval needed by a stone to reach the ground is 3 seconds and acceleration due to gravity is 10 m/s 2 . Determine the height of the building.

Time interval (t) = 3 seconds

Acceleration due to gravity (g) = 10 m.s -2

Wanted: Height of building (h)

Known: time interval (t) and acceleration due to gravity (g), wanted: height (h) so use the equation of free fall motion: h = ½ g t 2

h = ½ (10)(3)

h = 15 metes

8. A fruit free fall from its tree at the height of 12 m above the ground. If acceleration due to gravity is g = 10 m/s 2 and the friction of air ignored, then determine the height of the fruit above the ground after 1 second.

Height of tree (h) = 12 meters

Time interval (t) = 1 second

Wanted: The height of the fruit above the ground

After 1 second, fruit free fall as far as :

h = ½ g t 2 = ½ (10)(1) 2 = (5)(1) = 5 meters

The height of the fruit above the ground after 1 second :

12 meters – 5 meters = 7 meters

Answer : Free fall motion refers to the motion of an object under the influence of gravitational force only, with no other forces (like air resistance) acting on it.

Answer : All objects in free fall near the surface of the Earth experience a constant acceleration due to gravity, g , which is approximately 9.81 m/s 2 downward. This means that the object’s velocity increases by this amount for each second of free fall.

Answer : In the absence of air resistance, the mass of an object does not influence its free fall acceleration. All objects, regardless of their mass, will fall with the same acceleration due to gravity, g .

Answer : Astronauts inside the ISS appear to float not because there’s no gravity, but because both the astronauts and the ISS are in a continuous state of free fall around the Earth. They’re essentially falling at the same rate as the ISS, creating a sensation of weightlessness.

Answer : Mass is a measure of the amount of matter in an object and remains constant regardless of its location. Weight, on the other hand, is the force exerted on an object due to gravity. It varies depending on the gravitational field. During free fall, an object feels weightless because there’s no normal force acting on it, but its mass remains unchanged.

Answer : When an object is thrown upwards, it decelerates under the influence of gravity. Its velocity decreases until it becomes zero at its highest point. As it starts falling back down, it accelerates due to gravity, increasing its velocity in the downward direction.

Answer : Terminal velocity is the constant maximum velocity reached by a falling object when the downward force of gravity is balanced by the upward force of air resistance. At this point, the object no longer accelerates and continues to fall at a constant speed.

Answer : The time it takes for an object to reach the ground is proportional to the square root of the height from which it falls (assuming no air resistance). An object dropped from a greater height will take longer to reach the ground than one dropped from a shorter height.

Answer : As an object falls freely under gravity, its potential energy (relative to the ground) decreases. This decrease in potential energy is converted into kinetic energy, causing the object’s speed to increase.

• If two objects of different shapes but the same mass are dropped from the same height in a vacuum, which will hit the ground first?

Answer : In a vacuum, where there’s no air resistance, both objects will hit the ground at the same time. Their shape won’t matter because only gravity is acting on them, and their masses are the same.

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Mechanics (Essentials) - Class 11th

Course: mechanics (essentials) - class 11th   >   unit 4.

• Sign of gravity in free fall
• Free fall 1 body - solved example

Free fall - 2 body solved numerical

• Freefall: graphs and conceptual questions
• Free fall - total time up & down solved example
• Solving freefall problems using kinematic formulas
• Worked example: Free fall, object thrown up from a building
• Advanced: Freefall problems

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Video transcript

Unit 1B Workflow

• Big Ideas: Unit 1B
• Introduction: Kinematics
• Remarks: Significant Figures
• Video Lab: Measurement
• Presentation: Vectors and Scalars
• Virtual Activity: Working with Vectors
• Presentation: Unit Vectors and Vector Math
• Practice Problems: Vectors
• Presentation: Motion Studies
• Practice Problems: Motion Studies
• Presentation: Graphical Analysis of Motion
• Virtual Activity: Motion Graphs
• Practice Problems: Motion Graphs
• Presentation: Constant Acceleration Kinematics
• Practice Problems: Kinematics
• Presentation: Free-Fall Kinematics
• Demonstration: Free-Fall on the Moon
• Practice Problems: Free Fall Kinematics
• Challenge Problem: Free Fall Kinematics
• Lab Activity: Kinematics
• Review: Unit 1B
• Test: Unit 1B

Physics B Click here to see the unit menu Return to the home page to log out  

Practice Problems: Free Fall Click here to see the solutions.

1. A rock is dropped from a garage roof from rest.  The roof is 6.0 m from the ground. a. (easy) Determine how long it takes the rock to hit the ground. b. (easy) Determine the velocity of the rock as it hits the ground. c. (moderate) A second rock is projected straight upward from ground level at the moment the first rock was released.  This second rock had an initial upward velocity of +6.0 m/s.  How long will this second rock take to reach maximum height? d. (hard) At what time after release will the two rock have the same height?

2. (moderate) A projectile is shot upward at 189 m/s from a height of 20 m off the ground. How long will it take for the projectile to be at ground level?

3. (moderate) A hot-air balloon is hovering over a large public park. The operator of the balloon then makes the balloon begin to rise at a constant rate of 0.8 m/s. At some height from the ground the operator drops a rock from the basket. The rock take 10.3 seconds to hit a target on the ground. How high (above the ground) was the rock when it was released and what was its maximum height?

Please supplement these problems with those found in your companion text.

*The "AP" designation is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, products sold on this website.

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• School Guide
• Class 11 Syllabus
• Class 11 Revision Notes
• Maths Notes Class 11
• Physics Notes Class 11
• Chemistry Notes Class 11
• Biology Notes Class 11
• NCERT Solutions Class 11 Maths
• RD Sharma Solutions Class 11
• Math Formulas Class 11
• CBSE Class 11 Physics Notes

Chapter 1: Physical World

• What is Physics? Definition, History, Importance, Scope
• How is Physics related to Other Sciences?
• Fundamental Forces

Chapter 2: Units and Measurement

• System of Units
• Length Measurement
• Measurement of Area, Volume and Density
• Rounding Numbers
• Dimensional Analysis
• Significant Figures
• Errors in Measurement

Chapter 3: Motion in a Straight Line

• What is Motion?
• Distance and Displacement
• Speed and Velocity
• Acceleration
• Uniform Acceleration
• Sample Problems on Equation of Motion

Solving Problems Based on Free Fall

• Relative Motion
• Relative Motion in One Dimension
• Relative Motion in Two Dimension
• Calculating Stopping Distance and Reaction Time

Chapter 4: Motion in a Plane

• Scalar and Vector
• Vector Operations
• Product of Vectors
• Scalar Product of Vectors
• Dot and Cross Products on Vectors
• Position and Displacement Vectors
• Average Velocity
• Motion in Two Dimension
• Projectile Motion
• Uniform Circular Motion
• Centripetal Acceleration
• Motion in Three Dimensions

Chapter 5: Laws of Motion

• Contact and Non Contact Forces
• Inertia Meaning
• Law of Inertia
• What is Impulse?
• Solving Problems in Mechanics
• Linear Momentum of a System of Particles
• Newton's Second Law of Motion: Definition, Formula, Derivation, and Applications
• Laws of Conservation of Momentum
• What is Equilibrium? - Definition, Types, Laws, Effects
• Law of Action and Reaction
• Types of Friction - Definition, Static, Kinetic, Rolling and Fluid Friction
• Increasing and Reducing Friction
• Factors Affecting Friction
• Motion Along a Rough Inclined Plane
• Problems on Friction Formula
• Centripetal and Centrifugal Force
• Solved Examples on Dynamics of Circular Motion
• Dynamics of Circular Motion
• Motion in a Vertical Circle

Chapter 6: Work, Energy and Power

• Work Energy Theorem
• Practice Problems on Kinetic Energy
• Work Done by a Variable Force
• What is Potential Energy?
• Potential Energy of a Spring
• Practice Problems on Potential Energy
• Law of Conservation of Energy
• Difference Between Work and Energy
• Types of Collisions
• Collisions in One Dimension
• Collisions in Two Dimensions

Chapter 7: Systems of Particles and Rotational Motion

• Rigid Body - Definition, Rotation, Angular Velocity, Momentum
• Motion of a Rigid Body
• Centre of Mass
• Center of Mass of Different Objects
• Motion of Center of Mass
• Torque and Angular Momentum
• What are Couples? Definition, Moment of Couple, Applications
• What is the Principle of Moments?
• Centre of Gravity
• Moment of Inertia
• Kinematics of Rotational Motion
• Dynamics of Rotational Motion
• Angular Momentum in Case of Rotation About a Fixed Axis
• Rolling Motion
• Relation between Angular Velocity and Linear Velocity

Chapter 8: Gravitation

• Kepler's Laws of Planetary Motion
• Universal Law of Gravitation
• Factors affecting Acceleration due to Gravity
• Variation in Acceleration due to Gravity
• Potential Energy
• Escape Velocity
• Binding Energy of Satellites
• Weightlessness

Chapter 9: Mechanical Properties of Solids

• Elastic Behavior of Materials
• Elasticity and Plasticity
• Stress and Strain
• Hooke's Law
• Stress-Strain Curve
• Young's Modulus
• Shear Modulus and Bulk Modulus
• Poisson's Ratio
• Elastic Potential Energy
• Stress, Strain and Elastic Potential Energy

Chapter 10: Mechanical Properties of Fluids

• Fluid Pressure
• Pascal's Law
• Variation of Pressure With Depth
• How to calculate Atmospheric Pressure?
• Hydraulic Machines
• Streamline Flow
• Bernoulli's Principle
• Bernoulli's Equation
• What is Viscosity?
• Stoke's Law
• Reynolds Number
• Surface Tension

Chapter 11: Thermal Properties of Matter

• Difference between Heat and Temperature
• Temperature Scales
• Ideal Gas Law
• Thermal Expansion
• Heat Capacity
• Calorimetry
• Change of State of Matter
• Latent Heat
• Thermal Conduction
• Sample Problems on Heat Conduction
• What is Radiation - Types, Scource, Ionizing and Non-Ionizing Radiation
• Greenhouse Effect
• Newton's Law of Cooling

Chapter 12: Thermodynamics

• Thermodynamics
• Zeroth Law of Thermodynamics
• Heat, Internal Energy and Work
• First Law of Thermodynamics
• Specific Heat Capacity
• Thermodynamic State Variables and Equation of State
• Thermodynamic Processes
• Second Law of Thermodynamics
• Reversible and Irreversible Processes

Chapter 13: Kinetic Theory

• Behavior of Gas Molecules - Kinetic Theory, Boyle's Law, Charles's Law
• Molecular Nature of Matter - Definition, States, Types, Examples
• Kinetic Theory of Gases
• Mean Free Path - Definition, Formula, Derivation, Examples

Chapter 14: Oscillations

• Oscillatory and Periodic Motion
• Simple Harmonic Motion
• Force Law for Simple Harmonic Motion
• Displacement in Simple Harmonic Motion
• Velocity and Acceleration in Simple Harmonic Motion
• Energy in Simple Harmonic Motion
• Some Systems executing Simple Harmonic Motion

Chapter 15: Waves

• Introduction to Waves - Definition, Types, Properties
• Speed of a Travelling Wave
• Reflection of Waves
• Properties of Waves
• Principle of Superposition of Waves
• Energy in Wave Motion
• Doppler Effect - Definition, Formula, Examples

An object when in rest position has energy stored in it and when the object starts to move, it is said that the object is in motion. An object in motion has Kinetic energy, and it is not generated, it is obtained by simply converting the energy object had when it was in rest position. Motion can be defined in different ways, 1-D motion or one-dimensional motion is when the object moves in 1 co-ordinate only, for example, A boy riding his cycle in a straight line. 2-D or two-dimensional motion is when the object moves in 2 co-ordinates, mostly in both x and y co-ordinates, for example, children playing and running around in different directions. 3-D or three-dimensional motion is observed when an object moves in all three dimensions, for example, an airplane or a fighter jet moves in all three directions.

There are some special types of motion like Circular Motion, Parabolic motion, Free fall. However, these also come under the category of the ones mentioned above, that is, Circular motion is a 2- D motion, Parabolic motion is 2-D as well, Free fall is One dimensional. The most basic type of motion to learn and understand is 1-D motion. Let’s learn about 1-D motion in more detail,

Motion in a Straight Line

A body travelling in a straight line and in one direction has motion in a straight line. Examples can be a car moving in a straight line or an object free falling. The body is known to have uniform motion in a straight line if it travels equal distance in per unit time. For example, if a car travels 3 meters every second, it is covering equal distance every second, the speed of the body is 3m/sec. 

Uniform motion in a straight line

The other case can be having uniform acceleration, in this case, the body has constant acceleration, the speed changes uniformly.

The body/object is known to have non-uniform motion in a straight line if it travels unequal distances in equal intervals of time. For example, A car traveling 2 meters for the first second and 3 meters for the next second.

Non- uniform motion in a straight line

Free fall of an object is defined when the object is only under the influence of the force of gravity and no other force is acting upon it. In real life, an object when undergoes free fall, it also experiences other forces like air friction, etc. Hence, it will not be considered a Free Fall. Instances of Free fall can be the moon revolving around the sun, as it only experiences force under 

Equations of Motion for Free Fall

• First Equation of motion

Since, during Free fall, the initial velocity of the object is 0m/sec and the acceleration acted upon is the acceleration due to gravity (g= 9.8 m/sec2). the equation will look like,

u=0m/sec, a=g= 9.8m/sec 2

• Second Equation of motion

As mentioned above, the second equation of motion under free fall shall be,

S = ut+ 1/2(at 2 )

S = H, u=0m/sec, a= g= 9.8 m/sec 2

H = 1/2(gt 2 )

• Third Equation of motion

The third equation of motion under free fall,

v 2 = u 2 + 2aS

S= H, u=0m/sec, a= g= 9.8m/sec 2

Table to show the Final Velocities, distance covered, in every second of a Free Fall,

Sample Problems

Question 1: What happens when an object undergoes free fall?

When an object undergoes free fall, it starts with zero initial velocity, and keeps on increasing its velocity with a rate of 9.8 m/sec, hence, it experiences an acceleration of 9.8 m/sec 2 which is also known as acceleration due to gravity.

Question 2: What is the final velocity of the ball if it is dropped from a certain height and takes 10 seconds to reach the ground. The air resistance is not taken into account. [take g= 10 m/sec 2 ]

Applying first equation of motion under free fall, u= 0m/sec, S=H, a= g= 10m/sec 2 (given) v= u+ at v= gt v= 10× 10 v= 100m/sec

Question 3: Rohan dropped his ball from a height of 5 meters and then he ran downstairs really fast to see if he can catch the ball, he takes 1 minute to reach the ground. Will he be able to catch the ball, please note that the air friction is neglected.

Applying second equation of motion under free fall, S= ut+ 1/2(at 2 ) u=0 m/sec, a= g= 9.8m/sec 2 , S=H H= 1/2(gt 2 ) 5= 1/2(9.8× t 2 ) t 2 = 10/9.8 t= 1.01 second As it is very clear that Rohan takes 1 whole minute to reach the ground while the ball reaches the ground in 1.01 second. Hence, there is no way he will be able to catch the ball.

Question 4: If an object undergoes Free fall, what will be the velocity of the object on the 10th second?

Applying first equation of motion under free fall, v= gt v= 9.8× 10 v= 98m/sec Therefore, on the 10th second, the velocity of the object will be 98m/sec.

Question 5: Find the height from which a toy is dropped, if it goes in free fall state and the final velocity achieved by the toy is 20m/sec.

Applying third equation of motion, v 2 = u 2 + 2aS Under free fall, u= 0m/sec, a= g= 9.8m/sec 2 , S=H v 2 = 2gH 20 2 = 2× 9.8× H 400 = 19.6× H H= 20.40 meters.

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As Sony's mobile sales fall fast, I still don't know why Xperia exists — and that's the problem

What you need to know.

• An IDC analyst shared in a Bloomberg newsletter that sales for Sony's smartphone business dropped 40% last year.
• The drops are expected to continue as Sony's Xperia brand struggles to compete with the likes of Apple and Google.
• Sony is teasing the launch of a new Xperia phone during a May 15 announcement, and it is rumored to be targeting more of a mainstream audience. 

Sony is just days away from launching a brand-new Xperia phone, and this reveal sounds like it will have a big impact on what mobile offerings from the company will look like moving forward. Per rumors and leaks , the smartphone is expected to depart somewhat from Sony's typical strategy for handsets, which is to lean into the company's photography prowess and target the niche and premium markets. The strategy shift comes as Sony's mobile sales continue to take massive hits, according to data shared in a Bloomberg newsletter .

Masafumi Inbe, an analyst for IDC, spoke to Bloomberg about Sony's struggles with Xperia, which have stretched even to its home market in Japan. Apple has a powerful user base in Japan and found success in bringing the iPhone to that market. Now, part of Google's plan for growing Pixel is to target markets like Japan, too. 

Sony isn't a household name in smartphones anymore, and these developments further worsen its standing in the market. The company isn't one of the leaders in smartphone shipments, and it's going backward rather than forward. Part of the reason may be the direction of the Xperia brand, which is only appealing to a small number of people. 

However, switching things up and making the next Xperia 1 VI more mainstream might not be the answer. If Sony can't compete with the major brands while offering a unique smartphone, why would it compete with a device that feels like a carbon copy of all the others?

Sony's mobile sales are falling, fast

The global smartphone market has experienced frequent dips and peaks over the last few years, largely related to the fluctuating economy and world events. But Sony's mobile sales aren't just slightly down compared to past years; they're down significantly. According to Inbe, Sony's mobile sales dropped by 40% in 2023 compared to the previous year.

A similar cratering is expected this year, Inbe says, which doesn't bode well for the future of Sony's mobile offerings. If sales were to decline by a total of 80% over the course of two years, it would be disastrous. 

Inbe cites Sony's poor performance in Japan as one of the bigger reasons to sound the alarm. Sony sells about 1 million phones each year in Japan, Inbe says, where a total of 30 million smartphones are sold annually. 

Sony doesn't report smartphone sales figures individually in its financial reports, which makes it hard to get a complete picture of the business. However, it's clear that mobile devices aren't a core business for the company in 2024. Still, Sony hasn't shown any signs of backing out of the smartphone market.

"We recognize that our mobile communications business is not only a handset business, but also a business with important communication technology that contributes to the entire Sony Group," a Sony spokeswoman told Bloomberg.

"We believe that the communication technology we hone with smartphone technology is a necessary element to connect creators and customers and to provide entertainment regardless of time and place."

Something has to give, and if the leaks are correct, that something will be Xperia strategy. While it seems Xperia's current positioning toward the ultra-premium camera lover isn't working, it's unclear what will work this late in the smartphone race.

The Sony Xperia brand targets a market that barely exists

In theory, Sony's plan for the Xperia lineup sounds great. The company already makes industry-leading cameras, so why not bring that down to the smartphone? Although it's a solid idea, it isn't working. That's because the Sony Xperia brand only caters to people in the middle of two big markets. 

People who don't care much about standalone cameras will just grab any flagship smartphone, as the camera in a Google Pixel 8 Pro or an iPhone 15 Pro Max will get the job done in most situations. On the other hand, people that really care about cameras will get something like a DSLR or mirrorless camera on top of their smartphone. The number of people who need more than a smartphone but less than a dedicated shooter has proven to be small, if Xperia sales are any indication. 

As an anecdotal example, I'm the exact kind of buyer who should find the Sony Xperia line appealing. I have a mirrorless camera — from Sony, evidently — that I use for work and a bit of play. I'm not a photography buff, so I use my Sony a6400 pretty much stuck on automatic modes. Even with all that said, I wouldn't replace my smartphone or my mirrorless camera with an Xperia. For my fairly basic needs, a Sony Xperia smartphone would be a worse smartphone and a worse camera than what I currently use. So, what's the point? 

The Zenfone 11 Ultra proves making Xperia mainstream won't solve anything

Unfortunately, I don't have a solution for Sony. It feels like, in 2024, there are only a handful of brands that can really compete for the mainstream market. Although it appears Sony will try to do just that with the next Xperia, that might not be the best option.

We just saw Asus make a similar pivot with the Zenfone 11 Ultra , and the general consensus wasn't positive. Instead, it showed that Asus had given up a small but vibrant niche to become just like all the other flagship smartphones. And despite being one of the best Asus phones ever, the Zenfone 11 Ultra couldn't stand out.

If Xperia tries to compete on equal footing with Apple, Google, and others, it'll lose. The brand's only shot at this point is to stay the course and make niche, unique smartphones that actually cater to a tangible audience, and that's already turning into a losing battle.

Victim identity remains unknown after Ohio State University graduation death

Police in Columbus, Ohio, are investigating after someone fell to their death at the Ohio State University's main stadium Sunday afternoon.

Questions remain about what happened in the moments leading up to the fall, and authorities have not yet released the person's identity.

On Monday, the Franklin County coroner's office said that someone was pronounced dead at Ohio Stadium around 12:30 p.m. and that authorities were working to identify the body .

On Sunday afternoon, medics responded to the gates of the university's famed football arena, home of the Buckeyes football team, after police reported someone fell "off" the stadium, according to the Columbus Dispatch , part of the USA TODAY Network. In a statement Sunday, the university said the person had died after falling from the stands.

Photos from the scene Sunday show police and yellow tape near gate 30 of Ohio Stadium.

Ohio State University spokesperson Benjamin Johnson told USA TODAY early Monday that the school had no update on the investigation. Public information officers for the Columbus Police Department did not immediately respond to requests for comment Monday.

The Dispatch reported that the graduation ceremony continued uninterrupted after news of the fall spread throughout the crowd. Ceremony speakers did not mention the incident, according to the outlet.