Question

2. A boy stands in front of a cliff, on the other side of a river. He fires a gun and hears an echo after 6 seconds.
The boy then moves 170 m backwards and again fires the gun. He hears an echo after 7 seconds. Calculate :
(a) width of the river (b) speed of sound.​

Answer 1

Answer:

Explanation:

Solution,

Let us assume the speed of sound be v m/s.

And the width of the river be d₁

1st Case,

When the boy is at point A.

t = 6 seconds.

⇒ v = 2d₁/6

⇒ v = d₁/3 ...... (i)

2nd Case,

When the boy moves 170 m away at point B.

Distance = d₁ + 170 m

t = 7 seconds

⇒ v = 2(d₁ + 170)/7 ...... (ii)

Now, Solving Equation (i) and (ii), we have

⇒ 2(d₁ + 170)/7 = d₁/3

⇒ 2d₁ + 340/7 = d₁/3

By cross multiplication, we get

⇒ 7(d₁) = 3(2d₁ + 340)

⇒ 7d₁ = 6d₁ + 1020

⇒ 7d₁ - 6d₁ = 1020

⇒ d₁ = 1020

Hence, the width of the river is 1020 m.

By putting d₁'s value in eq (i), we get

⇒ v = d₁/3

⇒ v = 1020/3

⇒ v = 340

Hence, the speed of the sound is 340 m/s.

Answer 2

Answer:

here you go

hope it helped