Question

4. A man standing in front of a vertical cliff fires a gun. He hears the echo after 3 s. Then he moves
closer to the cliff by 82.5 m and fires the gun again. This time the echo reaches him in 25 s. Calculate
a) distance of cliff from initial position of man.​

Answer 1

Correct Question is :-

• A man standing in front of a vertical cliff fires a gun. He hears the echo after 3 s. 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 distance of cliff from initial position of man.

(Note :- As man reaches near to cliff, the time of echo reaches to him reduces not increases. Thats why time should be 2.5 seconds instead of 25 seconds)

$\large\underline{\bold{Solution-}}$

• Let the distance of the cliff from the initial position of the man at A be 'd' meter.

So,

• The distance traveled by sound in 3 sec = 2d meter.

So,

• Speed of sound, S is given by

$\sf \: Speed = \dfrac{Distance}{Time}$

$\bf \: \therefore \:\: Speed \:of \: sound, S = \dfrac{2d}{3} \: m \: per \: sec - - (1)$

Now,

• On moving closer to the cliff by a distance of 82.5 m,

• The distance travelled by sound in 2.5 sec = 2(d − 82.5) m

So,

Speed of sound, S is given by

$\sf \: Speed = \dfrac{Distance}{Time}$

$\bf \therefore \: Speed \: of \: sound, S = \dfrac{2(d - 82.5)}{2.5} \: m \: per \: sec - - (2)$

From equation (1) and equation (2), we get

$\sf \: \dfrac{2d}{3} = \dfrac{2(d - 82.5)}{2.5}$

$\sf \: \dfrac{2d}{3} = \dfrac{2d - 165}{2.5}$

$\sf \: 5d = 6d - 495$

$\therefore \: \boxed{ \bf{d \: = \: 495 \: m}}$

Additional Information :-

Echo

Echo is defined as a sound repeating by sound wave reflection, having a lasting or far reaching impact, or repeating, what someone else has said.

Necessary Conditions for the formation of an Echo

• (i) Time gap between the Sound must come back to the person after 0.1 second.

• (ii) For above condition, the reflecting surface must be at a minimum distance of 17.2m. It also depends on temperature.

Answer 2

Answer:

Correct Question is :-

A man standing in front of a vertical cliff fires a gun. He hears the echo after 3 s. 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 distance of cliff from initial position of man.

(Note :- As man reaches near to cliff, the time of echo reaches to him reduces not increases. Thats why time should be 2.5 seconds instead of 25 seconds)

$\large\underline{\bold{Solution-}}$

Let the distance of the cliff from the initial position of the man at A be 'd' meter.

So,

The distance traveled by sound in 3 sec = 2d meter.

So,

Speed of sound, S is given by

$\sf \: Speed = \dfrac{Distance}{Time}$

$\bf \: \therefore \:\: Speed \:of \: sound, S = \dfrac{2d}{3} \: m \: per \: sec - - (1)$

Now,

On moving closer to the cliff by a distance of 82.5 m,

The distance travelled by sound in 2.5 sec = 2(d − 82.5) m

So,

Speed of sound, S is given by

$\sf \: Speed = \dfrac{Distance}{Time}$

$\bf \therefore \: Speed \: of \: sound, S = \dfrac{2(d - 82.5)}{2.5} \: m \: per \: sec - - (2)$

From equation (1) and equation (2), we get

$\sf \: \dfrac{2d}{3} = \dfrac{2(d - 82.5)}{2.5}$

$\sf \: \dfrac{2d}{3} = \dfrac{2d - 165}{2.5}$

$\sf \: 5d = 6d - 495$

$\therefore \: \boxed{ \bf{d \: = \: 495 \: m}}$

Additional Information :-

Echo

Echo is defined as a sound repeating by sound wave reflection, having a lasting or far reaching impact, or repeating, what someone else has said.

Necessary Conditions for the formation of an Echo

(i) Time gap between the Sound must come back to the person after 0.1 second.

(ii) For above condition, the reflecting surface must be at a minimum distance of 17.2m. It also depends on temperature.