Question

A man standing in front of a vertical cliff fires a gun. He hears the echo after $3\text{s}$. On moving closer to the cliff by $82.5\text{m}$, he fires again and hears the echo after $2.5\text{s}$. Find :
$\left(a\right)$ the distance of the cliff from the initial position of the man, and $\left[2\text{marks}\right]$
$\left(b\right)$ the speed of sound. $\left[2\text{marks}\right]$

Answer 1

The figure above shows the two positions of the man in front of a cliff.Let the distance of cliff from the initial position of man be $d$ and the speed of sound be $v$
For the first echo, $t=\frac{2d}{v}=3\text{s}$ (given) ...$\left(i\right)$
On moving closer to the cliff by $82.5\text{m}$, the distance of cliff from the new position becomes $(d—82.5)\text{m}$,
then for second echo,
$t=\frac{2(d-82.5)}{v}=2.5\text{s}$ (given) ...$\left(ii\right)$
(a) Dividing eqn. $\left(i\right)$ by eqn. $\left(ii\right)$, we get
$\frac{d}{d-82.5}=\frac{3}{2.5}=\frac{6}{5}$
or $6d—495=5d$
or $d=495\text{m}$ $\left[2\text{marks}\right]$
(b) From eqn. $\left(i\right)$, $v=\frac{2d}{t}$
Substituting the value of $d=495\text{m}$, and $t=3\text{s}$
$v=\frac{2\times 495}{3}$
$⟹v=330\text{m/s}$ $\left[2\text{marks}\right]$