Posted by **Anonymous** on Sunday, October 2, 2011 at 6:05pm.

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

- Physics -
**bobpursley**, Sunday, October 2, 2011 at 6:40pm
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.

## Answer This Question

## Related Questions

- physics - A ball is dropped from a height of 20m and rebounds with a velocity ...
- physics - Two students are on a balcony that is 15 m above the ground. One ...
- Physics - A Stone Of Mass 1kg Is Dropped From Rest From A Window 20m Above The ...
- Physics - A ball is dropped from a height of 20 meter and rebounded with a ...
- Physics - A tennis ball is dropped from 1.80 m above the ground. It rebounds to ...
- physics - A ball is tossed so that it bounces off the ground, rises to height ...
- physics - A ball is tossed so that it bounces off the ground, rises to a height ...
- physics - a tennis ball is dropped from 1.30 meters above the ground. it ...
- physics - a .140 kg baseball is dropped from rest from a height of 2.2m above ...
- physics - a .140 kg baseball is dropped from rest from a height of 2.2m above ...

More Related Questions