Question

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`.

Answer 1

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)g t_(1)^(2)`
`500=0xxt_(1)+(1)/(2)xx10xxt_(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 = 1.47 s`
Therefore, the splash is heard at the top after time, t
Where,` t= t_(1) +t_(2) = 10 + 1.47 = 11.47 s`.