A stone dropped from the top of a tower of height 300 m high splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is $340 m s^{- 1}$ ? $(g = 9.8 m s^{- 2})$

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Solution

Height of the tower, s = 300 m
Initial velocity of the stone, u = 0
Acceleration, $a = g = 9.8 m / s^{2}$
Speed of sound in air = 340 m/s
The time $(t_{1})$ taken by the stone to strike the water in the pond can be calculated using the second equation of motion, as:
$s = u t_{1} + \frac{1}{2} g t_{1}^{2} 300 = 0 + \frac{1}{2} \times 9.8 \times t_{1}^{2} ∴ t_{1} = \sqrt{\frac{300 \times 2}{9.8}} = 7.82 s$
Time taken by the sound to reach the top of the tower, $t_{2} = \frac{300}{340} = 0.88 s$
Therefore, the time after which the splash is heard, $t = t_{1} + t_{2}$
=7.82+0.88=8.7 s

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