Question

A man stands in between two parallel cliffs and explodes a cracker. He hears the first echo after 0.6 s and the second echo after 2.4 s. Calculate the distance between the cliffs. [Speed of sound is 336 ms⁻¹]

Answer 1

Distance of the man from the nearer cliff \[ d_1 = \frac{v \times t}{2} \]

\[d_1 = \frac{336 \times 0.6}{2}\]

∴ d₁ = 100.8 m

Distance of the man from the farther cliff \[ d_2 = \frac{v \times t}{2} \]

\[d_2 = \frac{336 \times 2.4}{2}\]

∴ d₂ = 403.2 m

Total distance between the two cliffs = d₁ + d₂

= (100.8 + 403.2)

= 504 m