A ball is let fall from a height `h_(0)`. There are `n` collisions with the earth. If the velocity of rebound after `n` collision is `v_(n)` and the ball rises to a height `h_(n)` then coefficient of restitution `e` is given by

Options

A. `e^n = sqrt((h_n)/(h_0))`
B. `e^n = sqrt((h_0)/(h_n))`
C. `n e = sqrt((h_n)/(h_0))`
D. `sqrt(n e) = sqrt((h_n)/(h_0))`

Answer

A

Explanation

Let the ball fall from height h velocity of hit = `sqrt(2gh)`
`implies` Velocity of round = `esqrt(2gh) implies` it rebounds to height d`e^2 h`.
`implies` According to collision, `h_1 = e^2 h_0, h_2 = e^2 h_1 …., h_n = e^2h_(n - 1)`….. Multiple the n equation
`implies h_(n) = e^(2n)h_(0) implies e^(n) = sqrt((h_n)/(h_0))`.