Question

A man stands in between two walls and bursts a balloon. He
hears two successive echoes after 0.5 seconds and 2.5
seconds. The distance between the walls when the speed of
sound is 332 m/s, is
​

Answer 1

Answer:

The observer hears the echo after 0.5 sec of clapping.

Let the distance of the observer from the cliff be d

In the interval of 0.5 sec sound travels to the cliff from its origin(hands of the observer) and travels back to the ears of the observer .

The distance travelled by the sound = 2d

As, speed = distance travelled / time taken

So, distance = speed X Time taken

2d = 320*0.5

or,

d = 80 m

Similarly calculate for t = 2.5 second.

2d=320*2.5

d=400 m

So, the distance between walls is 400+80=480 m

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Answer 2

Answer:

Step-by-step explanation:

By the formula we know that 2d = v*t

First wall :-

Given:-

v = 332 m/s

t = 0.5 sec

d= ?

2d = v*t

d = (v*t)/2

d = (332*0.5)/2

d = 166*0.5

d = 83 m

Second wall:-

Given:-

v = 332 m/s

t = 2.5 sec

d = ?

2d= v*t

d = (v*t)/2

d = (332*2.5)/2

d = 166*2.5

d = 415

Hence, total distance = 83 m + 415 m = 498 m

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