Question

A man standing infront of a vertical cliff fires a gun. He hears the echo after 3S.

on moving closer to cliff by 82.5m, he fires again a and hears the echo after 2.5s.

find the distance of the cliff from the initial position of the man and also the speed

of sound.​

Answer 1

Explanation:

Since ECHO IS REFLECTION OF SOUND THUS THE SOUND WILL TRAVEL TWO TIMES BETWEEN MAN AND CLIFF.

NOW LET'S TAKE THE Distance BETWEEN CLIFF AND MAN BE x M.

NOW SPEED = DISTANCE / TIME

NOW SPEED OF SOUND IN FIRST CASE = 2x/3

NOW WHEN MAN COMES CLOSER TO CLIFF BY 82.5m.

THEN DISTANCE SOUND TRAVELS BECOMES

X - 82.5

IN THIS CASE SPEED OF SOUND

= 2(X-82.5) / 2.5

SINCE NOW SOUND COMES IN 2.5 SECONDS.

NOW AS SPEED OF SOUND REMAINS CONSTANT THUS WE CAN SAY SPEED OF LIGHT IN FIRST CASE = SPEED OF LIGHT IN SECOND CASE.

THAT IS :

2x = 2(x - 82.5)

3 2.5

5x = 6x - 495

x = 495.

NOW SINCE INITIAL POSITION BETWEEN CLIFF AND MAN WAS X.

THEREFORE INITIAL DISTANCE BETWEEN MAN AND CLIFF IS 495METRE.

NOW SPEED OF SOUND IN THIS CASE

= 2 × 495 / 3

= 330m/s.

THUS SPEED OF SOUND IN THIS CASE IS 330m/s.

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$\huge\mathfrak\red{ THANK YOU}$

BY SCIVIBHANSHU

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