Question

Speed after `n` rebounds ?

Answer 1

`h_n =(v_0^2 epsilon^(2n))/(2g)`

Explanation:

After `n` rebounds the ascending speed is

`v_n = v_0 epsilon ^n` where `v_0 = sqrt((g h)/2)` is the initial speed.

with that initial speed, using the mechanical energy conservation relationship

`1/2m v_n^2=m g h_n` or `h_n = v_n^2/(2g) = (v_0^2 epsilon^(2n))/(2g)`

then the equation for `h_n` is

`h_n =(v_0^2 epsilon^(2n))/(2g)`