Question

If a person drops a pebble in a well of $100\text{m}$ depth, after how much time (in seconds) will he be able to hear the sound of pebble hitting the surface of water?
Given:
Acceleration due to gravity, $g=10{\text{m/s}}^2$
Speed of sound in air, $v=343\text{m/s}$

• A. $\sqrt{20}+0.3\text{s}$
• B. $0.3\text{s}$
• C. $\sqrt{20}\text{s}$
• D. $20\text{s}$

Answer 1

The correct option is A $\sqrt{20}+0.3\text{s}$

The question can be divided into two parts, where in the first part, the time taken by the pebble to reach the surface of the water is calculated and in the second part, the time taken by the sound wave to reach the person is calculated.
Given:
Since the pebble is dropped the initial velocity, $u=0\text{m/s}$
Depth of the well, $d=100\text{m}$
Acceleration, $g=10{\text{m/s}}^2$
Speed of sound, $v=343\text{m/s}$

Let $t_1$ be the time taken by the pebble to hit the surface of water.
From the second equation of motion,
$d=u{t}_1+\frac{1}{2}g{t}_1^2$
$100=(0\times t_1)+(\frac{1}{2}\times (10)\times t_1^2)$
$100=5t_1^2$
$⟹t_1=\sqrt{20}\text{s}$

We know, $\text{time}=\frac{\text{distance}}{\text{speed}}$
Let $t_2$ be the time taken for the sound wave to reach the person after hitting the surface of water.
$t_2=\frac{d}{v}=\frac{100}{343}\approx 0.3\text{s}$

Time after which the person hears the sound of pebble hitting the surface of water is,
$t=t_1+t_2$
$=\sqrt{20}+0.3\text{s}$.