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=(v-u)/2 *t, 20=(v-3/4v)/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=30-40/-10, t=1s. Thus, d time interval btw the first & second bounce is t1-t2=4-1=3s