physics
posted by Edwin on .
A ball is dropped from a height of 20m and rebounds with a velocity which is 3/4 of the velocity with which it hit the ground. What is the time interval between the first and second bounces?

the acceleration as it's coming down is +10. Let the velocity at which it hit the ground be V. So, the velocity at which it rebounds will be 3/4V. Thus, 4rm v=u+gt. Where u is 0, v=10t, d time at which it lands is t=v/10. 4rm s=(vu)/2 *t, 20=(v3/4v)/2 *v/10. Thus, 400=v^2/4. V^2=1600, v=40. Now, d time for hitting d ground is t=40/10, t=4s. 4rm v=u+(g)t, the time for rebounce t=3040/10, t=1s. Thus, d time interval btw the first & second bounce is t1t2=41=3s