Question

A person standing between two vertical cliffs and 480 m from the nearest cliff shouts. He hears the first echo after 3s and the second echo 2s later. Calculate:

1)The speed of sound.

2) The distance of the other cliff from the person

Answer 1

1) Let d₁ be the distance of the nearest cliff and d₂ be the distance of the farther cliff.

The time for the first echo is t₁ = 3s

The first echo will be heard from the nearest cliff

The total distance traveled by sound before  reaching the person is 2d₁

We know that

Speed of sound  =  v = `(2d)/t = (2d_1)/t_1`

`v = (2xx480)/3`= 320 m/s

Hence, the speed of sound is 320 m/s.

2) The second echo is heard 2 s after the first one.

Hence, t₂ = 3 + 2 = 5s

Again the sound travels a total distance 2d₂ before reaching the person.

So, we get

`v = (2d_2)/t_2`

`:. d_2 = (vt_2)/2 = (320 xx 5)/2` = 800 m

Hence, the distance between the other cliff and the person is 800 m.