A ball at rest dropped from a height of 12m It loses 25% of its kinetic energy on striking the ground and bouncing back to a height h. Then the value of h is
A) 3m
B) 6m
C) 9m
D) 12m

Hint:
Here in this question we used the concept of law of conservation of energy in which we compare the potential energy with the kinetic energy to find the value of height “h”.

Complete step-by-step answer:
When the ball is at height of 12m then it will have its potential energy
Potential energy = mgH
Where, m = mass
g = acceleration due to gravity
H = height
Here H = 12m
g = 10m/s
On putting the value in the equation
So Potential energy = m $ \times $ (120)
And when it is dropped its, Potential energy is converted into kinetic Energy
So by law conservation of energy, we can say that
m $ \times $(120) = kinetic energy
After striking the ground it will lost its 25% of energy
So we can say that it’s initial Kinetic energy = 25%
And final Kinetic energy equal to the 75%
Final kinetic energy = $\dfrac{{75 \times m\left( {120} \right)}}{{100}}$
Or it can be also written as
Final Kinetic Energy = 90m
As the ball bounce back so it will attain a height h so at that position it will have its potential energy
Again using the law of conservation of energy
Final kinetic energy equal to the new potential energy due to height h
90m = mgh
So h = 9m. Option (C) is correct.

Note:
Law of conservation of energy states that energy can neither be created nor be destroyed in an isolated system which is on consideration.