Physics
posted by Anonymous .
A ball dropped from rest on to horizontal ground 20m between rebounds with 3/4 of the velocity with which it hits the ground find time that elapses between the first and second in part of the ball with the ground

So the KE of the rebound is (3/4)^2 of the original energy (height 20), so the ball goes up 9/16 * 20 meters.
time to go up that height = time to fall that height.
time to fall that height:
h= 1/2 g t^2 or t= sqrt2h/g
total time on bounce then is 2t or 2 sqrt(9/16*20/9.8)= 3/2 sqrt(20/9.8) seconds
check my thinking.