Question

A person stands in between two parallel cliffs which are 99 m apart. He fires a gun and hears two successive echoes after 0.2 s and 0.4 s. Calculate : (a) the distance of the person from the nearer cliff (b) speed of sound.

Answer 1

Let the distance of the person from the nearer cliff be d₁ and from the farther cliff be d₂.

\[ d_1 = \frac{v \times 0.2}{2} = 0.1 \ v \]

\[ d_2 = \frac{v \times 0.4}{2} = 0.2 \ v \]

d₁ + d₂ = 99 m

0.1 v + 0.2 v = 99

0.3 v = 99

\[ v = \frac{99}{0.3} \]

v = 330 ms⁻¹

Distance of the person from the nearer cliff d₁ = 0.1 × 330 = 33 m