A man is standing between two parallel cliffs | vedclass

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Question

(B) Let the distances of the man from the two cliffs be $d_1$ and $d_2$ respectively.The total distance traveled by the sound for the first echo is $2

Vedclass · Accepted Answer

$850$

Vedclass · Answer

(B) Let the distances of the man from the two cliffs be $d_1$ and $d_2$ respectively.The total distance traveled by the sound for the first echo is $2d_1 = v 	imes t_1$,where $v = 340 \, ms^{-1}$ and $t_1 = 1.5 \, s$.$d_1 = \frac{v 	imes t_1}{2} = \frac{340 	imes 1.5}{2} = 170 	imes 1.5 = 255 \, m$.The total distance traveled by the sound for the second echo is $2d_2 = v 	imes t_2$,where $t_2 = 3.5 \, s$.$d_2 = \frac{v 	imes t_2}{2} = \frac{340 	imes 3.5}{2} = 170 	imes 3.5 = 595 \, m$.The total distance between the cliffs is $D = d_1 + d_2 = 255 + 595 = 850 \, m$.

$A$ man is standing between two parallel cliffs and fires a gun. If he hears the 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 $= 340 \, ms^{-1}$)

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