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A ball is dropped from some height and after first collision with the ground if it reaches $\frac{3}{4}^{\text {th }}$ of its original height, then the $\%$ loss of its energy.
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The correct answer is:
$25$
Let a ball is dropped from height $\mathrm{h}$ then initial potential energy $U_i=m g h$
Final height $=\frac{3}{4} h$
then final potential energy, $U_f=\frac{3}{4} \mathrm{mgh}$
The percentage loss of its energy
$$
\left|\frac{U_f-U_j}{U_j}\right| \times 100=\left|\frac{\frac{3}{4} m g h-m g h}{m g h}\right| \times 100 \%
$$
$$
=\frac{1}{4} \times 100=25 \%
$$
Final height $=\frac{3}{4} h$
then final potential energy, $U_f=\frac{3}{4} \mathrm{mgh}$
The percentage loss of its energy
$$
\left|\frac{U_f-U_j}{U_j}\right| \times 100=\left|\frac{\frac{3}{4} m g h-m g h}{m g h}\right| \times 100 \%
$$
$$
=\frac{1}{4} \times 100=25 \%
$$
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