Question

A man standing in front of a vertical cliff fires a gun. He hears the echo after 3 seconds. On moving closer to the cliff by 82.5 m, he fires again. This time, he hears the echo after 2.5 seconds. Calculate:

1. the distance of the cliff from the initial position of the man.
2. the velocity of sound.

Answer 1

Let the initial distance of the man from the cliff be \(d\) m and the speed of sound be \(v\) m/s.

(a) For the first position:

\[ v=\frac{2d}{3} \]

After moving \(82.5\) m closer, distance from the cliff is:

\[ d-82.5 \]

For the second position:

\[ v=\frac{2(d-82.5)}{2.5} \]

Since the speed is the same,

\[ \frac{2d}{3}=\frac{2(d-82.5)}{2.5} \]

\[ 2.5d=3d-247.5 \]

\[ 0.5d=247.5 \]

\[{d=495\ \text{m}} \]

(b) Now,

\[ v=\frac{2\times495}{3} \]

\[ {v=330\ \text{m/s}} \]