Question

A man standing in front of a vertical cliff fires a gun. He hears the echo after $3s$. On moving closer to the cliff by $82.5m$, he fires again and hears the echo after $2.5s$. Find: (i) the distance of the cliff from the initial position of the man and (ii) the speed of sound.

Answer 1

Let the distance of the cliff from the initial position of the man be $d$ m.

So, the distance traveled by sound in $3sec=2dm$

So, speed of sound, $S=\frac{Distance}{Time}$

$S=\frac{2d}{3}.....\left(i\right)$

On moving closer to the cliff by a distance of 82.5 m, the distance$=2(d-82.5)m$

So,

$S=\frac{d}{t}$

$S=\frac{2(d-82.5)}{2.5}$

$=\frac{2d-2\times 82.5}{2.5}.....\left(ii\right)$

Therefore form equation (i) and (ii)

$\frac{2d}{3}=\frac{2d-2\times 82.5}{2.5}$

$5d=6d-495$

$d=495m$

Thus it is the distance of the cliff from the initial position of the man.

Now,

$S=\frac{2\times 495}{3}=330m/s$