A stone is dropped into a well, 44.1 metres deep. The sound of the splash is heard 3.13 seconds after the stone is dropped. Find the velocity of sound in air.

Explanation

For vertically downward motion of the stone,
Initial velocity, u=0 (being dropped)
Acceleration due to gravity, g = 9.8 m `s^(-2)`
Distance covered, h = 44.1 m
Time taken, t = ?
We have
`h=ut+1/2 "gt"^2`
`h=1/2 "gt"^2" "( :. u=0)`
`t=sqrt((2h)/(g))`
`t=sqrt((2xx44.1)/(9.8))=sqrt9=3s`
Time taken by the sound from the splash to travel from water surface to the edge of the well = 3.13 - 3 = 0.13 s
For upward motion of the sound waves
Velocity= `("Distance")/("Time")` i.e., `v=h/t`
`v=(44.1)/(0.13) = 339.23 m s^(-1)`
`:.` Velocity of sound = 339.23 m `s^(-1)`