Answer :
Answer:
The tennis ball hit the ground at a speed of approximately [tex]5.22\; {\rm m\cdot s^{-1}}[/tex].
The tennis ball bounced away from the ground at a speed of approximately [tex]4.36\; {\rm m\cdot s^{-1}}[/tex].
The average acceleration of the tennis ball during the contact was approximately [tex]766\; {\rm m\cdot s^{-2}}[/tex].
Explanation:
If an object of mass [tex]m[/tex] is travelling at a speed of [tex]v[/tex], the kinetic energy [tex]\text{KE}[/tex] of that object will be [tex]\text{KE} = (1/2)\, m\, v^{2}[/tex]. At a height of [tex]h[/tex] above the ground, the gravitational potential energy [tex]\text{GPE}[/tex] of that object will be [tex]\text{GPE} = m\, g\, h[/tex].
If the drag on the tennis ball is negligible, energy will be conserved:
[tex]\begin{aligned}& \text{GPE at initial height} \\ &= \text{KE right before hitting the ground}\end{aligned}[/tex].
In particular, if the tennis ball was released from a height of [tex]h[/tex], its speed [tex]v[/tex] right before landing will ensure that [tex]m\, g\, h = (1/2)\, m\, v^{2}[/tex]. Rearrange to obtain [tex]v = \sqrt{2\, g\, h}[/tex]. Since the tennis ball was released from a height of [tex]h = 1.39\; {\rm m}[/tex], the speed of the ball right before landing will be:
[tex]\begin{aligned}v &= \sqrt{2\, g\, h} \\ &= \sqrt{2\, (9.8\; {\rm m\cdot s^{-2}}) \, (1.39\; {\rm m})} \\ &\approx 5.22\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
The same reasoning applies when the tennis ball bounces off the ground:
[tex]\begin{aligned}& \text{KE right after leaving the ground} \\ &= \text{GPE at max height after bouncing upward}\end{aligned}[/tex].
If the tennis ball bounced off the ground at a speed of [tex]v[/tex], the height [tex]h[/tex] that this tennis ball will reach should also satisfy [tex]m\, g\, h = (1/2)\, m\, v^{2}[/tex]. Rearrange to obtain [tex]v = \sqrt{2\, g\, h}[/tex]. Since the tennis ball reached a height of [tex]0.968\; {\rm m}[/tex], it would have bounced off the ground at a speed of:
[tex]\begin{aligned}v &= \sqrt{2\, g\, h} \\ &= \sqrt{2\, (9.8\; {\rm m\cdot s^{-2}}) \, (0.968\; {\rm m})} \\ &\approx 4.36\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
Acceleration measures the rate of change in velocity. The velocity of this tennis ball was downward right before landing, at approximately [tex](-5.22\; {\rm m\cdot s^{-1}})[/tex] (the negative sign indicates the direction of velocity.) The velocity of this ball points upward after bouncing back, at approximately [tex]4.36\; {\rm m\cdot s^{-1}}[/tex].
Hence, the change in the velocity of this tennis ball will be:
[tex]\begin{aligned} \Delta v &= v_{1} - v_{0} \\ &\approx4.356\; {\rm m\cdot s^{-1}} - (-5.220\; {\rm m \cdot s^{-1}}) \\ &\approx 9.576\; {\rm m\cdot s^{-2}}\end{aligned}[/tex].
Since this change in velocity required a duration of [tex]\Delta t = 0.0125\; {\rm s}[/tex], the average acceleration of this tennis ball would have been:
[tex]\begin{aligned} a &= \frac{\Delta v}{\Delta t} \\ &\approx \frac{9.576\; {\rm m\cdot s^{-1}}}{0.0125\; {\rm s}} \\ &\approx 766\; {\rm m\cdot s^{-2}} \end{aligned}[/tex].
