Question

A ball is dropped into a well. The water level in the well is at a depth $h$ below the top. If the speed of sound is $v$, then after what time the splash of sound is heard?

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

Answer 1

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

The total time will include the time taken by ball to reach the bottom of the well and the time taken by sound in travelling the distance across well.

The time taken by ball to reach the bottom of well can be calculated by:

$h=\frac{1}{2}g{t}^2$

or, $t=\sqrt{\frac{2h}{g}}$

Time taken by sound to reach top from bottom of well is $t=\frac{h}{v}$

Hence total time is $T=\sqrt{\frac{2h}{g}}+\frac{h}{v}=h(\sqrt{\frac{2}{gh}}+\frac{1}{v})$