A ball is dropped from a heigth H. It rebounds from the ground a number of times. If coefficient is e, to what height does it go after nth rebounding ?

Explanation

Let `v_(0)=` velocity with which ball strikes the ground first time, after `v_(n)` is velocity after nth rebound.
`:. e=(v_(1))/(v_(0))=(v_(2))/(v_(1))=(v_(3))/(v_(2))=....=(v_(n))/(v_(n-1))`
`:. e=(v_(1))/(v_(0))xx(v_(2))/(v_(1))xx(v_(3))/(v_(2))=....=(v_(n))/(v_(n-1))=(v_(n))/(v_(0))` ...(i)
But `v_(0)=sqrt(2gh)` and `v_(n)=sqrt(2gH)`.
Here, H is the height to which the ball rises after nth rebounding.
From (i), `e^(n)=(v_(n))/(v_(0))=(sqrt(2gH))/(sqrt(2gh))=sqrt((H)/(h))`
`:. H=e^(2n)xxh`