Question

A body is dropped into a deep well. Sound of
splash is heard after 4.23 s. If the depth of the
well is 78.4m, the velocity of sound in air is
(nearly)
(1) 330 m/s
(2) 341 m/s
(3) 380 m/s
(4) 540 m/s​

Answer 1

HEY MATE ........

$4.23 = t1 + t2 \\ \\ s = ut + \frac{1}{2} a {t}^{2} \\ \\ a = g(9.8m {s}^{2} ) \\ \\ u = 0 \\ t = t1 \\ s = 78.4m \\ \\ 78.4 = 0 + \frac{1}{2} \times 9.8 \times {t}^{2} 1 \\ \\ \implies \frac{78.4}{4.9} = {t}^{2} 1 \\ \\ \implies \: t \: 1 = 4 \: sec \\ \\ \implies \: t2 = 4.23 - 4 \\ \\ = 0.23 |sec| \\ \\ v = \frac{78.4}{0.23s} \\ \implies \: v = 340.87m |s|$

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