A ball is released from rest relative to the elevator at a distance `h_1` above the floor. The speed of the elevator at the time of ball release is `v_0`. Determine the bounce height `h_2` relative to elevator of the ball (a) if `v_0` is constant and (b) if an upward elevator accleration `a=g/4` begins at the instant the ball is released. The coefficient of restitution for the impact is `e`.

Answer

A, B

Explanation

(a) `u_r=0`, `a_r=g`
`:. v_r=sqrt(2gh_1)`
After collision relative velocity
`v_r^()'=esqrt(2gh_1)`
and relative retardation is still g (downwards).
Hence,
`h_2=((v_r^()')^2)/(2g)=e^2h_1`
(b) `u_r=0`, `a_r=g+g/4=(5g)/(4)`
`:.` Just before collsion `v_r=sqrt(2((5g)/(4))h_1)`
Just after collision `v_r^()'=ev_r`.
Relative retardation is still `(5g)/(4)`.
Hence, `h_2=((v_r^()')^2)/(2((5g)/(4)))=e^2h_1`