A person is standing at the sea shore. An observer on the ship which is anchored in between a vertical cliff and the person on the shore fires a gun. The person on the shore hears two sounds, 2 seconds and 3 seconds after seeing the smoke of the fired gun. If the speed of sound in the air is 320 ms^-1 then calculate:

(i) the distance between the observer on the ship and the person on the shore.

(ii) the distance between the cliff and the observer on the ship.

Let d₁ be the distance between the observer on ship and person on shore. The first sound heard by the person at shore after t₁ = 2 s will be the direct sound of fire.

From relation v = $\dfrac{d}{t}$,

320 = $\dfrac{d_1}{2}$ or d₁ = 640 m.

(ii) The second sound heard by the person at shore will be the echo heard after reflection from the cliff. Since, sound takes 2 s to reach from the observer to the person, so the time taken by the echo of fire to reach the observer will be t₂ = 3 - 2 = 1 s.

Let d₂ be the distance between the observer and the cliff. Then $\dfrac{2d$2}{t2} = v

Or d₂ = $\dfrac{vt_2}{2}$ = $\dfrac{320 \times 1}{2}$ = 160 m.