A ball of mass `m` is released from rest relative to elevator at a height `h`, above the floor of the elevator. After making collision with the floor of the elevator it rebounces to height `h_2`. The coefficient of restitution for Collision is `e`. For this situation, mark the correct statement(s).

Options

A. If elavator is moving done with constant velocity `v_(0), then h_(2)=e^(2)h_(1)`
B. If elevator is moving down with constasnt velocity `v_(0)`, then `h_(2)=e^(2)h_(1)-v_(0)^(2)/(2g)`
C. if elevator is moving down with constant velocity `v_(0)`, then impulse imparted by floor of the elevator of the ball is `m(sqrt(2gh_(2))+sqrt(2gh_(1))+2v_(0))` is the upward direction.
D. If elevator is moving with constant acceleration of `g//4` in upward direction, then it is not possible to determine a relation between `h_(1)` and `h_(2)`from the given information.

Answer

A

Explanation

As coefficient of restitution equation is valid in all fames of reference. So in the framd of lift
`e=("velocity of separation")/("velocity of approach") v_(f)/v_(i)=(sqrt(2gh_(2)))/(sqrt(2gh_(1)))`………i
`h_(2)=e^(2)h_(1)`
Impulse delivered by the elavator to ball is
`J=vecv_(mv)_(f)-vecv_(mv)_(i)=msqrt(2gh_(2))+sqrt(2gh_(1))`
If the elevator is accelarating then replacing g with `g+g/4` we get `h_(2)=e^(2)h_(1)`