Question

Sunil is standing between two walls the wall closest to him is at a distance of 660 metre if he hears the first echo after 4 seconds and another after 2 seconds then find the velocity of sound in the air and find what is the distance between the two walls?

Answer 1

Distance between the closest wall and Sunil = 360m

Time after which he hears echo = 4s

Velocity of sound = 2(Distance/time)

                            = 360*2/4 

                            = 180m/s

Let distance between two walls be (x)m 

Distance of another wall from Sunil = (x - 360)m

Time after which he hears echo = 4s + 2s = 6s

Velocity of sound = 2(Distance)/time

                            = 2(x-360)/6

Hence,

180 = 2(x-360)/6 (Velocity of sound remains constant)

⇒1080 = 2(x-360)

⇒540 = x - 360

⇒x = 540 + 360

Distance between two walls=900m .

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Answer 2

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The sound first travels from Sunil to first wall and then travels back to him.

So total distance travelled by sound= 2*660=1320

Time taken by Sound= 4 Seconds

So speed Of sound= Distance traveled /Time taken=1320/4=330m/s

Distance traveled by second echo= 6*330 =1980 m

Distance of 2nd wall from Sunil= 1980/2=990 m

Distance between two walls= 990-660=330m