Question

A rock dropped down a well takes 1.8 s to hit the water. How far below the top of the well is the surface of the water

Answer 1

Answer: 15.87 m

From the equation of motion:

$s=ut+\frac{1}{2}at^2$

where, $s$ is the distance traveled, $u$ is the initial velocity, $a$ is the acceleration and $t$ is the time.

The rock free falls under gravity. Initial velocity, $u=0 m/s$, $a=g=9.8m/s^2$

It took $t=1.8 s$ for rock to hit the water.

Substitute the values in the given equation:

$\Rightarrow s=0+\frac{1}{2}9.8m/s^2\times(1.8s)^2=15.87 m$

Hence, the water is 15.87 m below the top level of the well.