A body at rest falls through a height ' \(h\) ' with velocity ' \(\mathrm{V}\) '. If it has to fall down further for its velocity to become three times, the distance travelled in that interval is

The body acquired velocity \(\mathrm{V}\) when it falls through a height \(\mathrm{h}\), starting from rest.
\(
\begin{aligned}
& \therefore \mathrm{V}^2=2 \mathrm{gh} \\
& \therefore \mathrm{h}=\frac{\mathrm{V}^2}{2 \mathrm{~g}}
\end{aligned}
\)
If it falls further and attains velocity \(3 \mathrm{~V}\) and if the total height through which it falls is h', then
\(
\begin{aligned}
& \left(3 \mathrm{~V}^2\right)=2 \mathrm{gh}^{\prime} \\
& \therefore 9 \mathrm{~V}^2=2 \mathrm{gh}^{\prime} \\
& \therefore \mathrm{h}^{\prime}=\frac{9 \mathrm{~V}^2}{2 \mathrm{~g}}=9 \mathrm{~h} \\
& \therefore \mathrm{h}^{\prime}-\mathrm{h}=9 \mathrm{~h}-\mathrm{h}=8 \mathrm{~h}
\end{aligned}
\)