Asked by halley

After falling from rest from a height of 34 m, a 0.47 kg ball rebounds upward, reaching a height of 24 m. If the contact between ball and ground lasted 1.8 ms, what average force was exerted on the ball?
Answered by bobpursley
the velocity of the ball at the ground was

vf^2=2gh

vf=sqrt 2gh.

but this velocity reversed at impact, so the change in velocity is
2sqrt2gh

force*time=2sqrt(2gh)
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