Q125 / 160 · AP EAMCET 2013 · Physics
A ball at rest is dropped from a height of 12 . It losses $25 \%$ of its kinetic energy on striking the ground and bounces back to a height ' $h$ '. Then value of ' $h$ ' is
- A$3 \mathrm{~m}$
- B$6 \mathrm{~m}$
- Correct answerC$9 \mathrm{~m}$
- D$12 \mathrm{~m}$
Solution
Energy of balls at rest, $K_1=m g h_1$ and $K_2=m g h_2$ percentage loss in $\mathrm{KE}=\frac{K_1-K_2}{K_1} \times 100$
$$
\begin{array}{rlrl}
& & \frac{25}{100} & =\left(\frac{12-h^2}{12}\right) \\
\Rightarrow \quad & \frac{25 \times 12}{100} & =12-h_2 \\
\Rightarrow & h_2 & =12-3=9 \mathrm{~m}
\end{array}
$$
Hence, the correct answer is option: 'C'.