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 .... $ m$ (Velocity of sound in air $= 340 ms^{-1}$)

Answer

(b) $2{d_1} + 2{d_2} = v \times {t_1} + v \times {t_2}$

==> $2({d_1} + {d_2}) = v({t_1} + {t_2})$

${d_1} + {d_2} = \frac{{v({t_1} + {t_2})}}{2}$

$=\frac{{340 \times (1.5 + 3.5)}}{2} = 850\,  m.$

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