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Question
A person standing between the two vertical cliffs produces a sound. Two successive echoes are heard at 4 s and 6 s. Calculate the distance between the cliffs.
(Speed of sound in air = 320 ms-1)
Solution
Let distance of 1 st cliff = d1m
t = 4 sec
Speed of sound in air V = 320 ms-1
∴ d1 = `(v xx t)/2`
= `(320 xx 4)/2`
= 640 m
For cliff II d2 = `(v xx t)/2`
= `(320 xx 6)/2`
= 960 m
∴ Total distance between two cliff is
d = d1 + d2
= 640 + 960
= 1600 m
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