Question

A hailstone falls from a height of `5` `km`. Ignoring air resistance, what is the hailstone's final velocity? The acceleration due to gravity is `9.8` `ms^-2`.

Answer 1

If we assume no air resistance the velocity is given by:

`v^2=u^2+2ad` and `u=0`, so, rearranging:

`v=sqrt(2ad)=sqrt(2*9.8*5000)=313` `ms^-1` or about `1127` `km^-1`

Explanation:

In order to answer this question it is necessary to assume that air resistance does not act. This is not a very realistic assumption.

Any real object falling has a 'terminal velocity' at which the air resistance acting upward balances the gravitational force acting downward. At that velocity, there is no further acceleration.

For a human skydiver this velocity is about `56` `ms^-1` or `200` `kmh^-1`. A hailstone is both smaller and lighter, so it is hard to guess its terminal velocity, but it would be less than `313` `ms^-1`.