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?

Answer 1

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)