A body falls on a surface of coefficient of restitution 0.6 from a height of \(1 \mathrm{~m}\). Then the body rebounds to a height of

As the body falls from a height
\(\begin{array}{ll}
& \mathrm{u}=0 \\
& \mathrm{v}^2=\mathrm{u}^2+2 \mathrm{gh} \\
\therefore \quad & \mathrm{v}^2=2 \times 9.8 \times 1=19.6 \\
\therefore \quad & \mathrm{v}=\sqrt{19.6} \mathrm{~m} / \mathrm{s}
\end{array}\)
Coefficient of restitution \(\mathrm{e}=\frac{\mathrm{v}}{\mathrm{u}}\)
\(\begin{aligned}
\mathrm{e} & =\frac{\text { Velocity after collision }\left(\mathrm{v}_{\mathrm{f}}\right)}{\text { Velocity before collision }\left(\mathrm{v}_{\mathrm{b}}\right)} \\
\therefore \quad \mathrm{v}_{\mathrm{f}} & =\mathrm{e} \times \mathrm{v}_{\mathrm{b}} \\
\mathrm{v}_{\mathrm{f}} & =0.6 \times \sqrt{19.6} \mathrm{~m} / \mathrm{s}
\end{aligned}\)
After the body rebounds,
\(\begin{aligned}
& v^2=u^2-2 g h \\
& \Rightarrow u^2=2 g h
\end{aligned}\)
\(\therefore \quad \mathrm{h}=\mathrm{u}^2 / 2 \mathrm{~g} \text {. }\)
Here, \(\mathrm{u}=\mathrm{v}_{\mathrm{f}}\)
\(\begin{aligned}
\therefore \quad \mathrm{h} & =\frac{(0.6 \times \sqrt{19.6})^2}{2 \times 9.8} \\
& =0.36 \mathrm{~m}
\end{aligned}\)