A man standing in front of a vertical cliff fires a gun. He hears the echo after 3 s. On moving closer to the cliff by 82.5 m, he fires again and hears the echo after 2.5 s. The distance of cliff from the initial position of man is

Let distance of cliff from the initial position of man be d m and speed of sound be Vms^-1
For the first echo $\mathrm{t}=\frac{2\mathrm{d}}{\mathrm{V}}=3\mathrm{s}\left(\text{ given }\right)\to \left(\mathrm{i}\right)$

On moving closer to the cliff by 82.5 m, the distance of cliff from the new position becomes (d-82.5) m, then for second echo
$\mathrm{t}=\frac{2(\mathrm{d}-82.5)}{\mathrm{V}}=2.5\mathrm{s}\left(\text{ given }\right)\to \left(\mathrm{ii}\right)$
Dividing eqn (i) by eqn (ii), we get
$\begin{array}{l}\frac{\mathrm{d}}{\mathrm{d}-82.5}=\frac{3}{2.5}=\frac{6}{5}\\ \text{ Or }6\mathrm{d}-82.5=5\mathrm{d}\\ \text{ or }\mathrm{d}=495\mathrm{m}\end{array}$