Chapter 2: Describing Movement in One Dimension, 1-D Linear Kinematics
2.7 Falling Objects: Human Movement in the Vertical Direction
Authors: Paul Peter Urone, Roger Hinrichs
Adapted by: Rob Pryce, Alix Blacklin
Learning Objectives
By the end of this section, you will be able to:
- Describe the effects of gravity on people and objects in motion.
- Describe the motion of people and objects that are in free fall.
- Calculate the position and velocity of people and objects in free fall.
Falling objects form an interesting class of motion problems. 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. By applying the kinematic principles developed in the previous section to falling objects, we can examine some interesting situations and learn much about gravity in the process.
Gravity
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.
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 will reach the ground after a hard 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, while 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. Additionally, for objects that are falling a short height and/or are very heavy relative to their size (such as a shot put), the assumption that air resistance is negligible (and therefore can be ignored) is a reasonable assumption.
The force of gravity causes objects to fall toward the center of Earth. The acceleration of free-falling objects is therefore called the acceleration due to gravity. The acceleration due to gravity is constant, which means we can apply the kinematics equations to any falling object where air resistance and friction are negligible. This opens a broad class of interesting situations to us. The acceleration due to gravity is so important that its magnitude is given its own symbol in biomechanics, [latex]g[/latex]. It is constant at any given location on Earth and has the average value
[latex]g=9\text{.}{\text{81 m/s}}^{2}\text{.}[/latex]
Although [latex]g[/latex] varies from [latex]9\text{.}{\text{78 m/s}}^{2}[/latex] to [latex]9\text{.}{\text{83 m/s}}^{2}[/latex], depending on latitude, altitude, underlying geological formations, and local topography, the average value of [latex]9\text{.}{\text{80 m/s}}^{2}[/latex] will be used in this text unless otherwise specified. The direction of the acceleration due to gravity is downward (towards the center of Earth). In fact, its direction defines what we call vertical. Note that whether the acceleration [latex]\bar{a}[/latex] in the kinematic equations has the value [latex]+\bar{a}[/latex] or [latex]-\bar{a}[/latex] depends on how we define our coordinate system. Therefore, if we define the upward direction as positive, then [latex]\bar{a} = -g = -\text{9.81 m/s}^2[/latex], and if we define the downward direction as positive, then [latex]\bar{a} = +g = +\text{9.81 m/s}^2[/latex]. As in the previous chapters, the difference is arbitrary, and sometimes gravity (or downwards) is defined as positive for convenience, depending on the situation. The important thing is that the same frame of reference (direction) is used throughout the problem.
One-Dimensional Human 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 that the velocity (if there is any) is vertical. If the object is dropped, we know the initial velocity is zero. Once the object has left contact with whatever held or threw it, the object is in free-fall. Under these circumstances, the motion is one-dimensional and has constant acceleration of magnitude [latex]g[/latex]. Now that we’re discussing motion in a different direction, we’ll represent vertical displacement with the symbol [latex]y[/latex] and continue to use [latex]x[/latex] for horizontal displacement. Similarly, the velocities in these examples refer to the vertical velocity. Sometimes, we’ll use [latex]v_{y}[/latex] and [latex]v_x[/latex] to refer to vertical and horizontal velocities, respectively, when working on problems with motion in both the horizontal and vertical directions. These situations are discussed in the next chapter.
[latex]v_f=v_i-\text{g}\Delta t[/latex]
[latex]y_f=y_i+v_i\Delta t +\frac{1}{2}g\Delta t^2[/latex]
Example 2.14
A person standing on the edge of a high cliff throws a ball straight up with an initial velocity of 13.0 m/s. The rock misses the edge of the cliff as it falls back to earth. Calculate the position and velocity of the ball 1.00 s, 2.00 s, and 3.00 s after it is thrown, neglecting the effects of air resistance.
Strategy
Draw a sketch.
We are asked to determine the position [latex]y[/latex] at various times. In this example it is reasonable to take the initial position [latex]y_i[/latex] to be zero. This problem involves one-dimensional motion in the vertical direction. We use plus and minus signs to indicate direction, with up being positive and down negative. Since up is positive, and the rock is thrown upward, the initial velocity must be positive too. The acceleration due to gravity is downward, so [latex]g[/latex] (or[latex]\bar{a}[/latex]) is negative. It is crucial that the initial velocity and the acceleration due to gravity have opposite signs. Opposite signs indicate that the acceleration due to gravity opposes the initial motion and will slow and eventually reverse it.
Since we are asked for values of position and velocity at three times, we will refer to these as [latex]y_{f1}[/latex] and [latex]v_{f1}[/latex]; [latex]y_{f2}[/latex] and [latex]v_{f2}[/latex]; and [latex]y_{f3}[/latex] and [latex]v_{f3}[/latex].
Solution for Position: [latex]y_{f1}[/latex]
1. Identify the knowns. We know that
[latex]y_i = 0[/latex];
[latex]v_i = \text{13.0m/s}[/latex];
[latex]g = \text{-9.81m/s}^2[/latex]; and
[latex]\Delta t = \text {1.00 s}[/latex].
2. Identify the best equation to use. We will use
[latex]y_f=y_i+v_i\Delta t +\frac{1}{2}g\Delta t^2[/latex]
because it includes only one unknown, [latex]y_f[/latex] (or [latex]y_{f1}[/latex], here), which is the value we want to find.
3. Plug in the known values and solve for [latex]y_{f1}[/latex]
Discussion
The rock is 8.10 m above its starting point at t = 1.00 s, since [latex]y_{f1} > y_i[/latex]. It could be moving up or down; the only way to tell is to calculate [latex]v_{f1}[/latex] and find out if it is positive or negative.
Solution for Velocity: [latex]v_{f1}[/latex]
1. Identify the knowns. We know that
[latex]y_i = 0[/latex];
[latex]v_i = \text{13.0m/s}[/latex];
[latex]g = \text{-9.81m/s}^2[/latex]; and
[latex]\Delta t = \text {1.00 s}[/latex].
We also know from the solution above that [latex]y_{f1} = \text {8.10m}[/latex].
2. Identify the best equation to use. The most straightforward is:
[latex]v_f=v_i-\text{g}\Delta t[/latex].
3. Plug in the knowns and solve.
[latex]v_{f1}={0}-\text{gt}=\text{13}\text{.}\text{0 m/s}-\left(9\text{.}{\text{81 m/s}}^{2}\right)\left(1\text{.}\text{00 s}\right)=3\text{.}\text{20 m/s}[/latex]
Discussion
The positive value for [latex]v_{f1}[/latex] means that the rock is still heading upward at [latex]t = \text{1.00s}[/latex]. However, it has slowed from its original 13.0 m/s, as expected.
Solution for Remaining Times
The procedures for calculating the position and velocity at [latex]t = \text{2.00s}[/latex] and [latex]t = \text{3.00s}[/latex] are the same as those above. The results are summarized in Table 2.1 and illustrated in Figure 2.39
Table 2.1 Kinematics of a Thrown Ball
| Time, t | Position, y | Velocity, v | Acceleration, a |
| 1.00 s | 8.10m | 3.20 m/s | -9.81 m/s2 |
| 2.00 s | 6.40m | -6.60 m/s | -9.81 m/s2 |
| 3.00 s | -5.10 m | -16.4 m/s | -9.81 m/s2 |
Graphing the data can help us understand it more clearly.
Discussion
The interpretation of these results is important. At 1.00 s the rock is above its starting point and heading upward, since [latex]y_{f1}[/latex] and [latex]v_{f1}[/latex] are both positive. At 2.00 s, the rock is still above its starting point, but the negative velocity means it is moving downward. At 3.00 s, both [latex]y_{f3}[/latex] and[latex]v_{f3}[/latex] are negative, meaning the rock is below its starting point and continuing to move downward.
Notice that when the rock is at its highest point (at 1.5 s), its velocity is zero, but its acceleration is still -9.81 m/s2. In fact, its acceleration is -9.81 m/s2 for the whole trip—while it is moving up and while it is moving down. Note that the values for [latex]y[/latex] are the positions (or displacements) of the rock, not the total distances traveled.
Finally, note that free-fall applies to upward motion as well as downward. Both have the same acceleration—the acceleration due to gravity, which remains constant the entire time. Astronauts training in the famous Vomit Comet, for example, experience free-fall while arcing up as well as down, as we will discuss in more detail later.
Making Connections: Take-Home Experiment—Reaction Time
A simple experiment can be done to determine your reaction time. Have a friend hold a ruler between your thumb and index finger, separated by about 1 cm. Note the mark on the ruler that is right between your fingers. Have your friend drop the ruler unexpectedly, and try to catch it between your two fingers. Note the new reading on the ruler. Assuming acceleration is that due to gravity, calculate your reaction time. How far would you travel in a car (moving at 30 m/s) if the time it took your foot to go from the gas pedal to the brake was twice this reaction time?
Example 2.15
What happens if the person on the cliff throws the ball straight down, instead of straight up? To explore this question, calculate the velocity of the ball when it is 5.10 m below the starting point, and has been thrown downward with an initial speed of 13.0 m/s.
Strategy
Draw a sketch.
Since up is positive, the final position of the ball will be negative because it finishes below the starting point at [latex]y_i = 0[/latex]. Similarly, the initial velocity is downward and therefore negative, as is the acceleration due to gravity. We expect the final velocity to be negative since the ball will continue to move downward.
Solution
1. Identify the knowns.
[latex]y_i = 0[/latex];
[latex]y_{f} = \text {-5.10m}[/latex];
[latex]v_i = \text{-13.0m/s}[/latex];
[latex]g = \text{-9.81m/s}^2[/latex].
2. Choose the kinematic equation that makes it easiest to solve the problem. The equation [latex]v_f^2=v_i^{2}+2g\left(\Delta y\right)[/latex] works well because the only unknown in it is [latex]v_f[/latex].
(We will plug in [latex]y_f[/latex] and [latex]y_i[/latex] in for [latex]\Delta y[/latex], where [latex]\Delta y = y_f - y_i[/latex])
3. Enter the known values
where we have retained extra significant figures because this is an intermediate result.
Taking the square root, and noting that a square root can be positive or negative, gives
The negative root is chosen to indicate that the ball is still heading down. Thus,
Discussion
Note that this is exactly the same velocity the ball had at this position when it was thrown straight upward with the same initial speed. (See Example 2.14 and FIgure 2.41).
This is not a coincidental result. Because we only consider the acceleration due to gravity in this problem, the speed of a falling object depends only on its initial speed and its vertical position relative to the starting point. For example, if the velocity of the ball is calculated at a height of 8.10 m above the starting point (using the method from Example 2.14) when the initial velocity is 13.0 m/s straight up, a result of ±3.20 m/s is obtained. Here both signs are meaningful; the positive value occurs when the ball is at 8.10 m and heading up, and the negative value occurs when the ball is at 8.10 m and heading back down. It has the same speed but the opposite direction.
Another way to look at it is this: In Example 2.14, the ball is thrown up with an initial velocity of 13.0 m/s. It rises and then falls back down. When its position is y = 0 on its way back down, its velocity is -13.0 m/s. That is, vertical projectiles will have the same speed on its way down as on its way up. We would then expect its velocity at a position of y = -5.10m to be the same whether we have thrown it upwards at +13.0 m/s or thrown it downwards at -13.0 m/s. The velocity of the ball on its way down from [latex]y=0[/latex] is the same whether we have thrown it up or down to start with, as long as the speed with which it was initially thrown is the same. (of course, these conditions apply only when air resistance is assumed to be negligible)
Example 2.16
The acceleration due to gravity on Earth differs slightly from place to place, depending on topography (e.g., whether you are on a hill or in a valley) and subsurface geology (whether there is dense rock like iron ore as opposed to light rock like salt beneath you.) Although it is rarely done in biomechanics, the precise acceleration due to gravity can be calculated from measurements of a dropped object. An object, usually a metal ball for which air resistance is negligible, is dropped and the time it takes to fall a known distance is measured. See, for example, Figure 2.42. Very precise results can be produced with this method if sufficient care is taken in measuring the distance fallen and the elapsed time.
Suppose the ball falls 1.0000 m in 0.45173 s. Assuming the ball is not affected by air resistance, what is the precise acceleration due to gravity at this location?
Strategy
Draw a sketch.
We need to solve for acceleration [latex]a[/latex]. Note that in this case, displacement is downward and therefore negative, as is acceleration.
Solution
1. Identify the knowns. [latex]y_i = 0[/latex]; [latex]y_f = \text{1.0m}[/latex]; [latex]\Delta t = 0.45173[/latex]; [latex]v_i = 0[/latex].
2. Choose the equation that allows you to solve for [latex]g[/latex] using the known values.
3. Substitute 0 for [latex]v_i[/latex] and rearrange the equation to solve for [latex]g[/latex]:
Solving for [latex]g[/latex] gives
4. Substitute known values yields
Discussion
The negative value for [latex]g[/latex] indicates that the gravitational acceleration is downward, as expected. We expect the value to be somewhere around the average value of 9.81 m/s2, so 9.8010 m/s2 makes sense. Since the data going into the calculation are relatively precise, this value for [latex]g[/latex] is more precise than the average value of 9.81 m/s2; it represents the local value for the acceleration due to gravity.
Check Your Understanding
A stunt person jumps off a building and falls 30.0 meters before they hit the safety mat. Assuming they fall freely (there is no air resistance), how long does it take them to hit the safety mat?
Solution
We know that
initial position [latex]y_i = 0[/latex],
final position [latex]y_f = \text{-30.0m}[/latex], and
[latex]g = \text{-9.81 m/s}^2[/latex].
We can then use the equation [latex]y_f=y_i+v_i\Delta t+\frac{1}{2}g\Delta t^2[/latex] to solve for [latex]\Delta t[/latex]. Inserting the known values, we obtain:
[latex]y_f = 0 + 0 + \frac{1}{2}\Delta t^2[/latex]
[latex]\Delta t^2 = \frac{2y_f}{g}[/latex]
[latex]\Delta t = \sqrt{\frac{2y_f}{g}} = \sqrt{\frac{2\left(\text{-30.0m}\right)}{\text{-9.81m/s}^2}} = \sqrt{\text{6.12s}^2} = \pm \text{2.47s}[/latex]
where we take the positive value as the physically relevant answer (time cannot have a negative value in this case). Thus, it takes about 2.5 seconds for the stunt person to hit the safety mat.
