A ball is shot vertically upward from the surface of another planet. A plot of y vs t for the ball is shown in Fig 2-36, where y is the height of the ball above its starting point and t=0 at the instant the ball is shot. The figure’s vertical scaling is set by ${\mathbf{y}}_{\mathbf{s}}\mathbf{=}\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{m}\mathbf{.}$. What are the magnitudes of (a) the free fall acceleration on the planet and (b) the initial velocity of the ball?

The graph of y(m) against t(s) with vertical scaling set by ${\mathrm{y}}_{\mathrm{s}}=30.0\mathrm{m}.$.

The situation of a body moving freely in any direction in the presence of gravity is referred to as free fall.During the free fall, the body is acted upon by acceleration due to gravity.

Here, assume that the downward direction is negative. Also, the time taken by the object to reach the ground would be half of the total time of the flight. From the figure, it can be seen that the total time of flight is 5 s. Therefore, the time taken by the object to reach the ground is half of the time of flight. Therefore, the time is 2.5 sec

$$\begin{gathered} ∆\mathrm{y}={\mathrm{V}}_{\mathrm{i}}\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2 \\ -25=0-0.5\mathrm{g}{(2.5)}^2 \\ \mathrm{g}=-8\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the free-fall acceleration is $-8\mathrm{m}/{\mathrm{s}}^2$

Applying kinematic equations of motions and gravity calculated in part a) we get,

$$\begin{gathered} ∆\mathrm{y}={\mathrm{V}}_{\mathrm{i}}\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2 \\ -25\mathrm{m}={\mathrm{V}}_{\mathrm{i}}(2.5\mathrm{s})-\frac{1}{2}\times 8\mathrm{m}/{\mathrm{s}}^2\times {(2.5\mathrm{s})}^2 \\ {\mathrm{V}}_{\mathrm{i}}=20\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the initial velocity is $20\mathrm{m}/\mathrm{s}$.