Question

2. A person stands in between two parallel cliffs which are 99 m apart. He fires a gun and hears
two successive echoes after 0.2 s and 0.4 s. Calculate : (a) the distance of the person from the nearer cliff (b) speed of sound.
[(a) 33 m (b) 330 ms")​

Answer 1

distance of nearer cliff = 33 m and velocity = 330 m/s

Explanation:

Let the velocity of sound is v

The distance of man from nearer cliff is d

Therefore from farther cliff it will be = ( 99 - d )

As the sound covers the distance two time

Thus  2 d = v x t

or  t = 2 d/v

here t is time for the echo heard by man

In first case  0.2 = 2 d/v                                 I

In second case 0.4 =2 ( 99 - d )/v               II

Dividing II by I , we  have

2 = ( 99 - d )/d   or  3 d = 99

d = 33 m

Thus the distance of nearer cliff is 33 m

From I the velocity of sound v = 2 x 33/0.2 = 330 m/s

Answer 2

Explanation:

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