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Question
A man standing in front of a vertical cliff fires a gun. He hears an echo after 35s. On moving close to the cliff by 82.5 m he fires again. This time he hears the echo after 2.55s. Calculate the speed of sound, distance, and the initial position of man from the cliff.
Solution
Given: t = 35, d = x m
t1 = 2.55s, d1 = (x − 82.5)m
v = `"2d"/"t"="2x"/3`.........(i)
v = `"2d"_1/"t"_1=(2"x"-165)/2.5` .........(ii)
Comparing (i) and (ii)
`(2"x")/3=(2"x"-165)/2.5`
or 5x = 6x − 165 × 3
x = 495m
Substituting this value in (i)
v = `(2xx495)/3` = 330 m/s.
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