Question

A ball at rest is dropped from a height of 12m.it loses 25% of it's kinetic energy on strucking the ground.find the height to which it bounces.

Answer 1

Answer:

Nine meters.

Explanation:

At 12 m the ball has a potential energy of m x g x 12 (m is mass of the ball in Kg) ( Potential energy is mgh).

The whole potential energy is converted to kinetic energy just before the ball touches the ground.

i.e KE = m x 10 x 12.

Since its elastic collision the direction of ball reversed.

After touching the ground the ball has KE 75% of initial KE .

i.e KE = 75/100 x m x 10 x 12

1/2 x m x v² = 3/4 x m x 10 x 12.

v² = 3² x 4 x 5

v = 6√5 m/s.

Let it rebounces to a height h.

Now PE at h must be equal to KE at ground.

m x g x h = 1/2 x m x 36 x 5

10 x h = 18 x 5

h = 9 m.

So it rebounces to height of 9 m.