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.
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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 = $\frac{d}{t}$,
320 = $\frac{d_1}{2}$ 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 $\frac{2d_2}{t_2}$ = v
Or d2 = $\frac{vt_2}{2}$ = $\frac{320 \times 1}{2}$ = 160 m.