A pebble is dropped freely in a well from it's top. It takes 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^-1, find: (i) the depth of water surface, and (ii) the time when echo is heard after the pebble is dropped.

(i) From the equation of motion,

h = ut + $\dfrac{1}{2}$ gt²

Given,

g = 10 m s^-2

t = 20 s

u = 0

Substituting the values in the formula, we get,

$h = (0 \times 20) + (\dfrac{1}{2} \times 10 \times 20^2) \\[0.5em] h = 0 + (5 \times 400) \\[0.5em] h = 5 \times 400 \\[0.5em] h = 2000 m \\[0.5em]$

Hence, the height of the well is 2000 m

(ii) As we know,

time (t) = $\dfrac{\text{distance}}{\text{speed}}$

or

t = $\dfrac{\text {depth of well}}{\text{speed of sound}}$

Substituting the values in the formula, we get,

$t = \dfrac{2000}{330} \\[0.5em] \Rightarrow t = 6.1 s \\[0.5em]$

As the pebble reaches the ground after 20 s, hence echo will be heard after (20 + 6.1) s = 26.1 s