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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
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$9 \mathrm{~m}$
Energy of balls at rest, $K_1=m g h_1$ and $K_2=m g h_2$ percentage loss in $\mathrm{KE}=\frac{K_1-K_2}{K_1} \times 100$
$$
\begin{array}{rlrl}
& & \frac{25}{100} & =\left(\frac{12-h^2}{12}\right) \\
\Rightarrow \quad & \frac{25 \times 12}{100} & =12-h_2 \\
\Rightarrow & h_2 & =12-3=9 \mathrm{~m}
\end{array}
$$
$$
\begin{array}{rlrl}
& & \frac{25}{100} & =\left(\frac{12-h^2}{12}\right) \\
\Rightarrow \quad & \frac{25 \times 12}{100} & =12-h_2 \\
\Rightarrow & h_2 & =12-3=9 \mathrm{~m}
\end{array}
$$
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