A body of mass 'm' falls from height 'h' on the ground. If 'e' be the coefficient of restitution between the body and the ground, find (i) the velocity with which it rises after the `n^("th")` collision with the ground (ii) The height upto which it rises after the `n^("th")` collision.

Explanation

(i) Let the velocity with which the body falls of the ground is `u_(1),` then `u_(1)=sqrt(2gh)`
The velocity with which the body rebounds after collision is `v_(1)`, then
velocity of separation `=e xx "velocity of approach"`
`or vec(v_(1))=-evec(u_(1)) or v_(1)=eu_(1)" (Negative sign indicates upward direction)"`
After rising to maximum height, the body will hit the ground with speed `v_(1)=eu_(1)`. Hence the speed with which the body will rise will be `v_(2)=ev_(1)=e^(2)u_(1)`
Similarly, `v_(3)=ev_(2)=e^(3)u_(1)`
Proceeding similarly, the speed with which the body will rise after `n^("th")` collision is `v_(n)=e^(n)u_(1)`
(ii) The K.E. of the body just before hitting the ground first time is `K=(1)/(2)mu_(1)^(2)=mgh`
If `h_(1)` be the height upto which the body rises after first collision then
`mgh_(1)=(1)/(2)mv_(1)^(2)=(1)/(2)m(eu_(1))^(2)=e^(2).(1)/(2) m u_(1)^(2)=e^(2).mgh`
`"or "h_(1)=e^(2)h`
`"or "h_(2)=e^(2)h_(1)`
`"or "h_(2)=e^(4)h`
Similarly, `" "h_(n)=e^(2n)h`