A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? Given, g = 10 m s^{−2} and speed of sound = 340 m s^{−1}.

#### Solution

Height of the tower, *s* = 500 m

Velocity of sound, *v* = 340 m s^{−1}

Acceleration due to gravity, g = 10 m s^{−2}

Initial velocity of the stone, *u* = 0 (since the stone is initially at rest)

Time taken by the stone to fall to the base of the tower, *t*_{1}

According to the second equation of motion:-

`s=ut_1+1/2"gt"_1^2`

`500=0xxt_1+1/2xx10xxt_1^2`

`t_1^2=100`

**t _{1} = 10 s**

Now, time taken by the sound to reach the top from the base of the tower, `t_2=500/340-1.47s`

Therefore, the splash is heard at the top after time, *t*

Where, **t = t _{1} + t_{2} = 10 + 1.47 = 11.47 s**