The vertical force equals the vertical momentum change divided by 0.0565 s.
The vertical momentum change is
0.0609 kg*(Vi*sin 22 + Vr*sin 17.7)
Vi and Vb are the incident and bounced speeds in m/s.
The horizontal force equals the horizontal momentum change divided by 0.0565 s
Concert the velocities to m/s before computing momentum change.
A tennis ball of mass 0.0609 kg is served. It strikes the ground with a velocity of 52.5 m/s (117 mi/h) at an angle of 22.0° below the horizontal. Just after the bounce it is moving at 50.5 m/s at an angle of 17.7° above the horizontal. If the interaction with the ground lasts 0.0565 s, what average force did the ground exert on the ball?
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