A man stands in a narrow, steep-sided valley. When he shouts, he hears two echoes, one after $1\text{ s}$…
A man stands in a narrow, steep-sided valley. When he shouts, he hears two echoes, one after $1\text{ s}$ and other after $2\text{ s}$. If the velocity of sound in air is $330\text{ m/s}$, the width of the valley is [MGIMS 2011]
$330\text{ m}$
$495\text{ m}$
$660\text{ m}$
$990\text{ m}$
Solution
Let $s_1$ and $s_2$ be the distances of man from either side of valley, then $2s_1 = vt_1$ and $2s_2 = vt_2$
$\Rightarrow 2(s_1 + s_2) = v(t_1 + t_2)$
$\Rightarrow s_1 + s_2 = \frac{v(t_1 + t_2)}{2}$
$\therefore d = s_1 + s_2 = \frac{330 \times (1 + 2)}{2}$
$\Rightarrow d = \frac{330 \times 3}{2} = 495\text{ m}$