Question

A ball is dropped from a height of 20 mon a floor for which e - 1/2. The height attained by the ball
after the second collision​

Answer 1

Answer Hope it helps

Answer 2

Solution-
Initially it reaches ground with a velocity of
2gh
​
=20m/s
Let the velocity of separation after first collision be x
And the velocity of collision after second collision be y,
Then as per equation of e,
x/20=1/2 and y/x = 1/2

Solving we get y=5m/s

So height attained by ball after second collision=y^2/2g=1.25m