Question

A stone is dropped into a deep well and is heard to hit the water 3.41s after being dropped. Determine the depth of the well

Answer 1

56.9 meters assuming g=9.8m/s^2 and neglecting the propagation time of the sound from the water surface up the well.

Answer 2

The time from dropping until hearing the sound can be divided into 2 periods: the first is from when you drop the stone to when it reaches the end, the second is from when the stone reaches the end to when you hear the sound. We got:\[s=\frac{ g t _{1} ^ {2} }{ 2 }\rightarrow t _{1}=\sqrt{\frac{ 2s }{ g }}\] \[s=v _{sound}t _{2}\rightarrow t _{2}=\frac{ s }{ v _{sound} }\] \[t _{1}+t _{2}=3.41\rightarrow \sqrt{\frac{ 2s }{ g }}+\frac{ s }{ v _{sound} }=3.41\] Depend on what those constant, g and velocity of sound, are, you can find out the depth of the well.