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:

In order to determine the value of height "h," we employed the notion of the law of conservation of energy in this question. Specifically, we compared the potential energy with the kinetic energy.
The ball will have its potential energy when it is 12 meters in height.
Potential energy = mgH
Where, m = mass
g = acceleration due to gravity
H = height

Here $\mathrm{H}=12 \mathrm{m}$
$g=10 \mathrm{m}/\mathrm{s}$
Regarding entering the value into the equation
So Potential energy $=\mathrm{m}\times \left(120\right)$
Additionally, potential energy is transformed into kinetic energy when it is dropped. As a result, according to the law of energy conservation,
$m\times \left(120\right)=$ kinetic energy
We may say that its initial kinetic energy was 25% and that its final kinetic energy was equivalent to 75% because it lost 25% of its energy after impacting the ground.
Final kinetic energy $=\frac{75\times m\left(120\right)}{100}$

Or It can also be expressed as
Final Kinetic Energy = 90m
The ball will reach a height of h when it returns, and at that point, it will have all of its potential energy.
Using the principle of energy conservation once more
Final kinetic energy = new potential energy as a result of height h

$90m = mgh$

So, $h=9m$

Hence, the correct answer is option C.