Question
A tennis ball of mass $0.060 \mathrm{kg}$ is served. It strikes the ground with a velocity of $54 \mathrm{m} / \mathrm{s}(120 \mathrm{mi} / \mathrm{h})$ at an angle of $22^{\circ}$ below the horizontal. Just after the bounce it is moving at $53 \mathrm{m} / \mathrm{s}$ at an angle of $18^{\circ}$ above the horizontal. If the interaction with the ground lasts $0.065 \mathrm{s}$, what average force did the ground exert on the ball?
Step 1
This force acts in the Y direction. Therefore, we use the impulse equation $F \Delta t = m \Delta v_y$. Show more…
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