Question

A man standing in front of a vertical cliff fires a gun and hears the echo after $4$ seconds. He moves $150m$ towards the cliff and again fires the gun. This time, he hears the echo after $3.6$ seconds. Find the distance of the cliff from his first position.

Answer 1

Given:
Time to hear first echo, $t_1=4s$
Time to hear second echo, $t_2=3.6s$
Distance between first and second source locations, $s=150m$

Let the distance of cliff from the first location be $d$
Let the velocity of sound be $v$

Since echo is a reflection of the original sound, it will travel distance equal to twice the distance between man and cliff i.e., $2d$.
From definition of speed,
$v=\frac{2d}{t_1}$ ...........(i)

Now, after travelling 150 m towards the cliff, distance travelled by sound, $d_2=2(d-150)=2d-300$
Again from definition of speed,
$v=\frac{2d-300}{t_2}$ ................(ii)

Equating the value of $v$ from equations (i) and (ii), we get:
$\frac{2d-300}{t_1}=\frac{2d}{t_2}$
$\frac{2d-300}{3.6}=\frac{2d}{4}$

Solving the above equation, we get
$d=1500m$