Answer:
24 m/s
Step-by-step explanation:
Given that the height of the ball from the elevator, h=20 m
The acceleration due to gravity, g=10 m/

The initial velocity of the ball is zero, so, by using the equation of the motion
the velocity of the ball, v, when reached the elevator,

m/s
The velocity of the ball just before the collision is 20 m/s in the downward direction.
The velocity of the lift just before the collision is 2 m/s in the upward direction.
As the ball and the lift moves in the opposite direction, so the speed of approach
m/s.
Assuming that there is negligible change in the speed of the lift, so the velocity of the lift after the collision is 2 m/s
Let the velocity of the ball after the collision is
m/s in the direction of the velocity of the lift.
So, the speed of the separation

As the collision is elastic, so the coefficient of restitution,



m/s
Hence, the ball rebounds with a velocity of 24 m/s.