#### 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 340 m s^{–1}? (g= 9.8 m s^{–2})

#### Solution 1

Here, h = 300 m, g = 9.8 ms^{-2 } and velocity of sound, v = 340 ms^{-1 }Let t_{1}be the time taken by the stone to reach at the surface of pond.

Then using `s = ut + 1/2 at^2 1/2 at^2 => h = 0 xx t + 1/2 "gt"_1^2`

`:. t_1 = sqrt((2xx300)/9.8) = 7.82 s`

Also if `t_2` is the time taken by the sound to reach at a height h, then

`t_2 = h/v = 300/340 = 0.88 s`

:. Total time after which sound of splash is heard = `t_1 + t_2`

= 7.82 + 0.88 = 8.7 s

#### Solution 2

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

`300 = 0 + 1/2 xx 9.8 xx t_1^2`

`:. t_1 =sqrt((300xx2)/9.8) = 7.82 s`

Time taken by the sound to reach the top of the tower, `t_2 = 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