Question

A man standing in front of a vertical cliff fires a gun. He hears the echo after 3s. Then he moves closer to the cliff by 82.5 m and fires the gun again. This time the echo reaches him in 2.5 s
calculate:

a) Distance from cliff from initial position of the man.
b) velocity of sound
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Answer 1

Answer:

A) 495m

B) 330m/s

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Complete solution :-

a) Let the cliff be at a distance d from the man, then the time t after which the first eco is heard.

t = 2d/v -------> 1.)

Here, v is the velocity of sound.

b) In the second case, the time Y will be given by

Y = 2(d-82.5)/v -------> 2.)

By Dividing (2) By (1) We get :-

Y/t = d - 82.5/d

2.5/3 = 1- 82.5/d (by putting given values)

By solving d = 82.5×6 = 495m

Now,

v = 2d/t

v = 2× 495/3

v = 330 m per s.

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~ Nikhra♥