A pebble is dropped freely in a well from its top. It takes the $20 s$ for the pebble to reach the water surface in the well. Taking g = $10 m / s^{2}$ and speed of sound $= 330 m / s .$ find the time when the sound is heard after the pebble is dropped.

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Solution

Given: $t = 20 s, g = 10 m / s^{2}$
Assume ‘h’ to be the depth of the well,
We know from the equation of motion,
$H = u t + \frac{1}{2} g t^{2}$
$= 0 + \frac{1}{2} (10) (20)^{2}$
$= 2000 m$

$Finding the time when the sound is heard after the pebble is dropped - ------------------------------------------------------------------------ - :$
Given: speed of sound $= 330 m / s$
$T i m e t a k e n = \frac{d e p t h}{s p e e d}$
$= \frac{2000}{330} = 6.1 s$
For the pebble to reach the water surface, it takes $20 s,$
Hence the total time taken when the sound is heard after the pebble is dropped
$= 20 + 6.1 = 21.6 s$

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A pebble is dropped freely in a well from its top. It takes the 20s for the pebble to reach the water surface in the well. Taking g = 10m/s2 and speed of sound =330m/s. find the time when the sound is heard after the pebble is dropped.

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