Question

A stone dropped from the top of a tower of height 300 m 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)

Answer 1

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 (t₁ ) taken by the stone to strike the water in the pond can be calculated using the second equation of motion, as:

\(s=ut_1+\frac{1}{2}gt_1^2\)

\(300=0+\frac{1}{2}×9.8×t_1^2\)

\(∴ t_1=\sqrt\frac{300×2}{9.8}=7.82\,s\)

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