Question

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 $340m{s}^{-1}$ ? $(g=9.8m{s}^{-2})$

Answer 1

Height of the tower, s = 300 m
Initial velocity of the stone, u = 0
Acceleration, $a=g=9.8m/s^2$
Speed of sound in air = 340 m/s
The time $\left(t_1\right)$ taken by the stone to strike the water in the pond can be calculated using the second equation of motion, as:
$s=u{t}_1+\frac{1}{2}g{t}_1^{2}300=0+\frac{1}{2}\times 9.8\times t_1^2\therefore t_1=\sqrt{\frac{300\times 2}{9.8}}=7.82s$
Time taken by the sound to reach the top of the tower, $t_2=\frac{300}{340}=0.88s$
Therefore, the time after which the splash is heard, $t=t_1+t_2$
=7.82+0.88=8.7 s