Question
A man fires a bullet standing between two cliffs. First echo is heard after $3$ seconds and second echo is heard after $ 5$ seconds. If the velocity of sound is $330 m/s$, then the distance between the cliffs is .... $m$
✓
Answer
(b) $2({d_1} + {d_2}) = v({t_1} + {t_2})$==>${d_1} + {d_2} = \frac{{330 \times (3 + 5)}}{2} = 1320$ $m$
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