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.

Check Your Understanding

A chunk of ice breaks off a glacier and falls 30.0 meters before it hits the water. Assuming it falls freely (there is no air resistance), how long does it take to hit the water?


Solution

We know that initial position y_{0}=0, final position y=-30.0 \mathrm{~m}, and a=-g=-9.80 \mathrm{~m} / \mathrm{s}^{2} . We can then use the equation y=y_{0}+v_{0} t+\frac{1}{2} a t^{2} to solve for t . Inserting a=-g, we obtain

\begin{aligned}&y=0+0-\frac{1}{2} g t^{2} \\ &t^{2}=\frac{2 y}{-g} \\ &t \quad=\pm \sqrt{\frac{2 y}{-g}}=\pm \sqrt{\frac{2(-30.0 \mathrm{~m})}{-9.80 \mathrm{~m} / \mathrm{s}^{2}}}=\pm \sqrt{6.12 \mathrm{~s}^{2}}=2.47 \mathrm{~s} \approx
    2.5 \mathrm{~s} \end{aligned}

where we take the positive value as the physically relevant answer. Thus, it takes about 2.5 seconds for the piece of ice to hit the water.