Question

a ball is dropped from a height of 12 meters if the ball loses 25% of its Kinetic Energy on striking the ground what is the height to which it bounces

Answer 1

Before fall:
Let velocity of ball be v1 when it reaches ground after 12 metre fall. Then using equation of motion
Initial velocity (u1) of ball = 0m/s
Then using equation of motion: v² = u² + 2as
where symbols have there usual meaning
we get,
(v1)² = (u1)² + 2gs (g: acceleration due to gravity)
or (v1)² = 0² + 2g(12)
or (v1)² = 24g (m/s)²
Kinetic energy (K.E.) will be:
K.E. = $\frac{1}{2}$m(v1)² (m: mass of ball)
or K.E. = $\frac{1}{2}$m(24g) Joules

Now, ball loses 25% kinetic energy after drop. So it bounces back with 75% Kinetic energy. Therefore, maximum velocity it reaches after bounces be v2. So, kinetic energy (k2) of ball after bounce will be:
k2 = 75% of K.E.
or $\frac{1}{2}$m(v2)² = $\frac{75}{100}$($\frac{1}{2}$m(v1)²)
or (v2)² = (0.75)(24g) (m/s)²

At maximum height (h), the velocity of ball will be zero. Using equation of motion

v² = u² + 2as
we get,
(0)² = (v2)² + 2gh

or (v2)² = -2gh
or (0.75)(24g) = -2gh
or h = -9m (negative sign tells that displacement is in upward direction) ans

The height achieved is 9m in upward direction.