A ball of mass 0.1 kg, initially at rest, is dropped from height of 1 m. Ball hits the ground and bounces off the ground. Upon impact with the ground, its velocity reduces by 20%. The height (in m) to which the ball will rise is:

Answer 1

The ratio of KE after the bounce to before the bounce is

1/2 m(.8v)^2
-----------------
     1/2 mv^2
= .8^2
= .64

so, the ball will bounce .64 as high as the original height