# 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.

May 9, 2016

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

${v}^{2} = {u}^{2} + 2 a d$ and $u = 0$, so, rearranging:

$v = \sqrt{2 a d} = \sqrt{2 \cdot 9.8 \cdot 5000} = 313$ $m {s}^{-} 1$ or about $1127$ $k {m}^{-} 1$

#### 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 $56$ $m {s}^{-} 1$ or $200$ $k m {h}^{-} 1$. A hailstone is both smaller and lighter, so it is hard to guess its terminal velocity, but it would be less than $313$ $m {s}^{-} 1$.