A stone is dropped into a well. If the depth of water below the top be `h` and velocity of sound is `v` then the splash in water is heard after `T` sec. Then:

Options

A. ` T=2 h//v`
B. T= sqrt (2 h)/g + h /v`
C. `T= sqrt (2 h) /v + h/g`
D. ` T= sqrt ( h/(2g) + (2h)/v`

Answer

B

Explanation

Let ` t_1` be the tme taken by the stone to fall through a vetical distance `h` and ` t_1` be the time taken by the sound to travel distance `h`.
Since sound moves with a constant velocity,
:. ` t_2 = h/v`
` Taking downward motion, ` t-1 = sqrt (2 h)/g` ,
Total time ` - t_1 +t-2 = sqrt ( 2h)/g + h/v` .