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+(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+(1)/(2)9.8m/s^2*(1.8s)^2=15.87 m`

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