A stone is dropped into a well `45 m` deep. The sound of the splash of the splash is heard `3.13 s` after the stone is dropped. Find the speed of sound in air. Take `g = 10 m//s^2`.

Explanation

Here, depth of the well, `h = 45 m`
initial velocity of the stone, `u = 0, g = 10 m//s^2`
time taken by the sound of splash to be heard, `t_1 = 3.13 s`
If `t_2` is the time taken by the stone to reach water level in the well, then form
`h = ut_2 + (1)/(2) g t_2^2` or `45 = (1)/(2) xx 10 xx t_2^2`
or `t_2^2 = 9` or `t_2 = 3 s`
Clearly, time taken by sound to reach the top of the well, i.e., `t = t_1 = 3.13 - 3 s = 0.13 s`
Speed of sound `= ("distance (h) travelled by sound")/("time taken (t)")`
i.e., `v = (45)/(0.13) = 346 m//s`.