Question

A man is standing between two parallel cliffs and fires a gun. If he hears first and second echoes after 1.5 s and 3.5 s respectively, the distance between the cliffs is (Velocity of sound in air = $340m{s}^{-1}$)

• A. 1190 m
• B. 850 m
• C. 595 m
• D. 510 m

Answer 1

The correct option is B 850 m

Let the distance from man and both the cliffs be $d_1$ and $d_2$ respectively


$2d_1+2d_2=v\times t_1+v\times t_2$
$\Rightarrow 2(d_1+d_2)=v(t_1+t_2)$
$d_1+d_2=\frac{v(t_1+t_2)}{2}=\frac{340\times (1.5+3.5)}{2}=850m.$