Question

Suppose you drop a rock into a dark well and, using precision equipment, you measure the time for the sound of a splash to return. Neglecting the time required for sound to travel up the well, calculate the distance to the water if the sound returns in 2. 00 s.

Answer 1

Approximately
`19.6\; {\rm m}` (assuming that
`g = 9.81\; {\rm m\cdot s^(-2)}` and that air resistance is negligible.)

Step-by-step explanation:

Assume that the air resistance on the rock is negligible. During the descent, the acceleration
`a` of the rock will be constant:
`a = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}`.

It is given that the descent took
`t = 2.00\; {\rm s}`. Let
`x` denote the displacement (change in position) of the stone. Apply the SUVAT equation
`x = (1/2)\, a\, t^(2)` to find displacement
`x\!`:

`\begin{aligned}x &= (1)/(2)\, a\, t^(2) \\ &= (1)/(2)\, ((-9.81)\; {\rm m\cdot s^(-2)})\, (2.00\; {\rm s})^(2) \\ &\approx (-19.6)\; {\rm m}\end{aligned}`.

Note that
`x` is negative since the water is below the initial position of the rock. Therefore, the distance to the water will be approximately
`19.6\; {\rm m}`.