Question

A ball at rest is dropped into a well. The water is at a depth $h$ from the surface. If the speed of sound is $c$, then the time after which the splash is heard will be

• A. $h[\sqrt{\frac{2}{gh}}+\frac{1}{c}]$
• B. $h[\sqrt{\frac{2}{gh}}-\frac{1}{c}]$
• C. $h[\frac{2}{g}+\frac{1}{c}]$
• D. $h[\frac{2}{g}-\frac{1}{c}]$

Answer 1

The correct option is A $h[\sqrt{\frac{2}{gh}}+\frac{1}{c}]$


As the ball is released into well, let the time required for it to reach the surface of the water be $t_1$.

Using 2nd equation of kinematics:
$h=\frac{1}{2}g{t}_1^2$
$\Rightarrow t_1=\sqrt{\frac{2h}{g}}$

When the balls hits the water surface after time $t_1$, a splash of sound is created which travels back to the observer in time $t_2$.
Here, $h=c{t}_2$
$t_2=\frac{h}{c}$

So total time $\left(t\right)$ after which the splash is heard is:
$t=t_1+t_2=\sqrt{\frac{2h}{g}}+\frac{h}{c}$
$t=h(\sqrt{\frac{2}{gh}}+\frac{1}{c})$