As a hailstone falls under gravity in still air, its mass increases. At time \(t\) the mass of the hailstone is \(m\). The hailstone is modelled as a uniform sphere of radius \(r\) such that
$$\frac{dr}{dt} = kr,$$
where \(k\) is a positive constant.
- Show that \(\frac{dm}{dt} = 3km\). [2]
Assuming that there is no air resistance,
- show that the speed \(v\) of the hailstone at time \(t\) satisfies
$$\frac{dv}{dt} = g - 3kv.$$ [4]
Given that the speed of the hailstone at time \(t = 0\) is \(u\),
- find an expression for \(v\) in terms of \(t\). [5]
- Hence show that the speed of the hailstone approaches the limiting value \(\frac{g}{3k}\). [1]