Question

A 95 kg student slides down a waterslide that is 13 m high. How fast is he going at the bottom?

Answer 1

The speed is `v\approx 16.12m/s (\sqrt{260} m/s)`

Explanation:

The initial energy is just the potential one:

`PE=mgh`

At the bottom, there’s only kinetic energy:

`KE=1/2mv^2`

If there are no dissipative forces, then the energy is conserved. Therefore:

`KE=PE`

`1/2mv^2=mgh`

`\implies v=\sqrt{2gh}`

`v=\sqrt{20N/kg*13m}=\sqrt{260}m/s\approx 16.12m/s`

You don’t even need the mass of the student. :-)