# a person standing between two vertical cliffs and 640 m away from the nearest cliff, shouted. he heard the first echo after 4 s and the second echo 3 s later. the distance between the cliffs is

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## A person standing between two vertical cliffs and 640 m away from the nearest cliff shouted. He heard the first echo after 4 seconds and the second echo 3 seconds later. Calculate the distance between the cliffs.A. 176 mB. 1700 mC. 1760 mD. 1800 m

A person standing between two vertical cliffs and 640 m away from the nearest cliff shouted. He heard the first echo after 4 seconds and the second echo 3 seconds later. Calculate the distance between the cliffs.A. 176 mB. 1700 mC. 1760 mD. 1800 m

Byju's Answer Standard IX Physics Reflection of Sound A person stan... Question

A person standing between two vertical cliffs and 640m away from the nearest cliff shouted. He heard the first echo after 4 seconds and the second echo 3 seconds later. Calculate the distance between the cliffs.

A 176 m B 1700 m C 1760 m D 1800 m Open in App Solution

The correct option is **C**. 1760 m

Let the person be standing 640 metres away from Cliff A. It is given that the person shouted and he heard the echo after 4 seconds. When the person will shout, the sound waves will go from the person to Cliff A and return to him again. This means the sound waves travel a distance of 640 metres two times. Therefore the total distance travelled by the sound is

d = 640 × 2 = 1280 m .

This distance is travelled by sound in 4 seconds. So the speed of the sound is

v = 1280 4 = 320 m / s .

This speed of sound is constant.

Let the second echo strike the other cliff, say Cliff B. The time taken by the echo to return to the person is 4 + 3 = 7 seconds.

Since the speed of sound is 320 m/s, the total distance travelled by the sound waves is

d ′ = 320 × 7 = 2240 m .

But this is the total distance travelled by the sound waves, which will be two times the actual distance between the cliff and the person.

Therefore, the distance between the cliff and the person will be 2240/2 = 1120 m. So, we obtain two equations —

Distance between the person and Cliff A = 640 m.

Distance between the person and cliff B = 1120 m.

Therefore the distance between the two cliffs will be 640+1120 = 1760 m.

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SIMILAR QUESTIONS

**Q.**

A man standing in front of a vertical cliff fires a gun and hears the echo after

4 seconds. He moves 150 m

towards the cliff and again fires the gun. This time, he hears the echo after

3.6

seconds. Find the distance of the cliff from his first position.

EXPLORE MORE Reflection of Sound Standard IX Physics

## A person standing between two vertical cliffs and 320 m away from the nearest cliff, produces sound. He hears the first echo after 4 s and the second echo 1.5 s later. Calculate the distance between the cliffs.

Click here👆to get an answer to your question ✍️ ** 1 C) A person standing between the two vertical cliffs and 640m away from the nearest cliff, produces sound. He hears the first echo after 4s and the second echo 3s later. Calculate: i) the speed of sound in air and ii) the distance between the cliffs. (4)

Question

** 1 C) A person standing between the two vertical cliffs and 640m away from the nearest cliff, produces sound. He hears the first echo after 4s and the second echo 3s later. Calculate: i) the speed of sound in air and ii) the distance between the cliffs. (4)

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## A Person Standing Between Two Vertical Cliffs and is 640 M Away from the Nearest Cliff. He Shouted and Heard the First Echo After 45s and the Second Echo After Further 3s. Calculate the Speed

A Person Standing Between Two Vertical Cliffs and is 640 M Away from the Nearest Cliff. He Shouted and Heard the First Echo After 45s and the Second Echo After Further 3s. Calculate the Speed

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A person standing between two vertical cliffs and is 640 m away from the nearest cliff. He shouted and heard the first echo after 45s and the second echo after further 3s. Calculate the speed of sound in air and the distance between the cliffs.

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### SOLUTION

**Given:**d1 = 640m; t1 = 4s

t2 = 4 + 3 = 7

First echo is heard from the nearest cliff.

Total distance travelled = 2d = 2 × 640 = 1280 m

v = d t 2d1t1=12804 = 320 m/s

Second echo is heard from the first cliff

v = d t 2d2t2 ⇒ 320 ms−1 = 2d 2d27 d2 = 320×72 = 1120 m

hence, the distance between two cliff

d1 + d2 = 640 + 1120 = 1760 m.

Concept: Sound (Numerical)

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Chapter 6: Echoes and Vibrations of Sound - Long Numerical

Q 3 Q 2 Q 4

### APPEARS IN

ICSE Class 10 Physics

Chapter 6 Echoes and Vibrations of Sound

Long Numerical | Q 3

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Guys, does anyone know the answer?