Kinematic Equations for Objects in Free Fall

As you read, pay attention to the relevant equations in the box Kinematics Equations for Objects in Free Fall where Acceleration = −g

$v_{f}=v_{i}+a\Delta t$

$y_{f}=y_{i}+v_{i}t+\frac{1}{2}a\Delta t^{2}$

$v^{2}=v\tfrac{2}{i}+2a(y_{f}-y_{i})$

Note that because the motion is free fall, a is simply replaced with $-g$ (here, $g$ is the acceleration due to gravity, $g=9.80\ \mathrm{m/s}^{2}$ ) and the direction of motion is the $y$ direction, rather than the $x$ direction. When calculating the position and velocity of an object in freefall, we need to consider two different conditions. First, the object can be thrown up as it enters freefall. For example, you could throw a baseball up and watch it fall back down.

Complete the steps in Example 2.14. After you review the solution, pay attention to the graphs in Figure 2.40. You can throw an object directly downward as it enters freefall, such as when you throw a baseball directly down from a second-floor window.

Then, complete the steps in Example 2.15. Notice that Figure 2.42 compares what is happening in Example 2.14 and Example 2.15. It is important to understand the difference between an object that is thrown up and enters free fall, versus an object that is directly thrown down. We can often use experimental data to calculate constants, such as $g$ .

In Example 2.16, we determine the acceleration due to gravity constant ( $g$ ) from experimental data.

Example 2.15 Calculating Velocity of a Falling Object: A Rock Thrown Down

What happens if the person on the cliff throws the rock straight down, instead of straight up? To explore this question, calculate the velocity of the rock 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.

Figure 2.41

Since up is positive, the final position of the rock will be negative because it finishes below the starting point at $y_{0}=0$ . 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 rock will continue to move downward.

Solution

1. Identify the knowns. $y_{0}=0 ; y_{1}=-5.10 \mathrm{~m} ; v_{0}=-13.0 \mathrm{~m} / \mathrm{s} ; a=-g=-9.80 \mathrm{~m} / \mathrm{s}^{2}$ .

2. Choose the kinematic equation that makes it easiest to solve the problem. The equation $v^{2}=v_{0}^{2}+2 a\left(y-y_{0}\right)$ works well because the only unknown in it is $v$ . (We will plug $y_{1}$ in for $y$ ).

3. Enter the known values

$v^{2}=(-13.0 \mathrm{~m} / \mathrm{s})^{2}+2\left(-9.80 \mathrm{~m} / \mathrm{s}^{2}\right)(-5.10 \mathrm{~m}-0 \mathrm{~m})=268.96 \mathrm{~m}^{2} / \mathrm{s}^{2},$

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

$v=\pm 16.4 \mathrm{~m} / \mathrm{s}.$

The negative root is chosen to indicate that the rock is still heading down. Thus,

$v=-16.4 \mathrm{~m} / \mathrm{s}.$

Discussion

Note that this is exactly the same velocity the rock had at this position when it was thrown straight upward with the same initial speed. (See Example 2.14 and Figure 2.42 (a)). 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 rock 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 $\pm 3.20 \mathrm{~m} / \mathrm{s}$ is obtained. Here both signs are meaningful; the positive value occurs when the rock is at 8.10 m and heading up, and the negative value occurs when the rock is at 8.10 m and heading back down. It has the same speed but the opposite direction.

Figure 2.42 (a) A person throws a rock straight up, as explored in Example 2.14. The arrows are velocity vectors at 0, 1.00, 2.00, and 3.00 s. (b) A person throws a rock straight down from a cliff with the same initial speed as before, as in Example 2.15. Note that at the same distance below the point of release, the rock has the same velocity in both cases.

Another way to look at it is this: In Example 2.14, the rock is thrown up with an initial velocity of $13.0 \mathrm{~m} / \mathrm{s}$ . It rises and then falls back down. When its position is $y=0$ on its way back down, its velocity is $-13.0 \mathrm{~m} / \mathrm{s}$ .

That is, it has the same speed on its way down as on its way up. We would then expect its velocity at a position of $y=-5.10 \mathrm{~m}$ to be the same whether we have thrown it upwards at $+13.0 \mathrm{~m} / \mathrm{s}$ or thrown it downwards at $-13.0 \mathrm{~m} / \mathrm{s}$ . The velocity of the rock on its way down from $y=0$ 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.