A person standing between two parallel cliffs | vedclass

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(B) Let the person be at a distance $d_1$ from the first cliff and $d_2$ from the second cliff.The time taken for the first echo is $t_1 = 1$ s. The s

Vedclass · Accepted Answer

$660$

Vedclass · Answer

(B) Let the person be at a distance $d_1$ from the first cliff and $d_2$ from the second cliff.The time taken for the first echo is $t_1 = 1$ s. The sound travels a distance of $2d_1$ to return to the person.So, $2d_1 = v 	imes t_1 = 330 	imes 1 = 330$ m.$d_1 = 165$ m.The time taken for the second echo is $t_2 = 3$ s. The sound travels a distance of $2d_2$ to return to the person.So, $2d_2 = v 	imes t_2 = 330 	imes 3 = 990$ m.$d_2 = 495$ m.The total distance between the two cliffs is $D = d_1 + d_2 = 165 + 495 = 660$ m.

$A$ person standing between two parallel cliffs fires a gun and hears two echoes, the first echo after $1$ second and the second echo after $3$ seconds. The distance between the two cliffs is (The velocity of sound = $330$ m/s). (in $m$)

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