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

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Answer 1 · 2018-04-19T15:40:41

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

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. :-)