Question

A man stands in between two cliffs X and Y, such that he is at a distance of 66 m from X. When he blows a whistle he hears the first echo after 0.4 s and the second echo after 1.2 s. Calculate : (a) speed of sound (b) distance of cliff Y from the man.

Answer 1

\[ \text{(a) Speed of sound } v = \frac{2d}{t}\]

\[= \frac{2 \times 66}{0.4}\]

= 330 ms⁻¹

\[ \text{(b) Distance of the man from cliff Y, } d = \frac{v \times t}{2} \]

\[ = 330 \times \frac{1.2}{2}\]

= 198 m