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

Explanation

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)xx9.8 xx t_(1)^(2)`
`therefore 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)=088 s`
Therefore , the time after which the splash is heard, `t=t_(1)+t_(2)`
`=0.72+0.88 =8.7 s`