Question

A man stands in front of a cliff and fires a gun. He hears an echo after 4 seconds. On moving closer to the cliff by 41.25 m and again firing a gun, he hears an echo after 3.75 s. Calculate : (a) speed of sound and (b) distance of the man from the cliff to his initial position.

Answer 1

Let x m be the initial distance of the man from the cliff and v the speed of sound.

\[ \text{Case (i) } t = \frac{2d}{v} \]

\[ \therefore \ 4 = \frac{2x}{v} \quad \text{or} \quad v = \frac{x}{2} \qquad \text{...(i)} \]

\[ \text{Case (ii) } t = \frac{2d}{v} \]

\[ 3.75 = \frac{2(x - 41.25)}{v} \]

\[ \text{or} \quad v = \frac{2x - 82.5}{3.75} \qquad \text{...(ii)} \]

\[ \text{Comparing (i) and (ii)} \]

\[ \frac{x}{2} = \frac{2x - 82.5}{3.75}\]

⇒ 3.75 = 4x − 165

∴ x = 660 m

\[ \text{Substituting the value of } x \text{ in (i)} \]

\[ v = \frac{660}{2} = 330 \ \mathrm{ms^{-1}} \]