Derive an expression for the height attained by a freely falling body after 'n' number of rebounds from the floor.

Explanation

Let a small ball be dropped from a height 'h' on a horizontal smooth plate . Let it rebounds to a height `'h_(1)'` .

Velocity with which it strikes the plate `u_(1)=sqrt(2gh)`
Velocity with which it leaves the plate `v_(1)=sqrt(2gh_(1))`
The velocity of plate before and after collision is zero i.e ., `u_(2)=0,v_(2)=0`
Coefficient of restitution
`e=("relative velocity of separatio ")/("relative velocity of approach")`
`=(sqrt(2gh_(2)-0))/(sqrt(2gh_(1)-0))thereforee=(sqrth_(1))/(h)" or " h_(1)=e^(2)h`
For 2nd rebound it goes to a height
`h_(2)=e^(2)h_(1)=e^(2)e^(2)h=e^(4)h`
For 3rd rebound it goes to a height
`h_(3)=e^(2)h_(2)=e^(2)e^(4)h=e^(6)h`
For nth rebound height attained
`h_(n)=e^(2n)h.`