A student stands 330 m from a tall cliff which is 990 m from another tall cliff.

The student fires a pistol and hears some echoes. If the speed of sound between the cliffs is 330 m/s,

(a) What is the time taken by first echo to reach the student?

(b) Find the time interval between the first echo and the second echo.

(a) Given,

D₁ = 330 m

D₂ = 660 m

V = 330 ms^-1

t₁ = ?

t₂ = ?

As we know,

$\text {Time interval (t)}_1 = \dfrac{\text{Total distance travelled (2d)}}{\text {Speed of sound (V)}} \\[0.5em]$

$\text{t}$1 = \dfrac{2 \times 330 }{330} \\[0.5em] \Rightarrow \text{t}1 = 2 \text{s}

Hence, time taken by first echo to reach the student = 2 s

(b) time taken by second echo to reach the student :

$\text{t}$2 = \dfrac{2 \times 660 }{330} \\[0.5em] \Rightarrow \text{t}2 = 4 \text{s}

t₂ - t₁ = 4 - 2 = 2 s

Hence, time interval between two echoes is 2 s