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?
Answers
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)
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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