7. 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 $$\frac { \mathrm { d } r } { \mathrm {~d} t } = k r$$ where \(k\) is a positive constant.

1. Show that \(\frac { \mathrm { d } m } { \mathrm {~d} t } = 3 \mathrm {~km}\). Assuming that there is no air resistance,
2. show that the speed \(v\) of the hailstone at time \(t\) satisfies $$\frac { \mathrm { d } v } { \mathrm {~d} t } = g - 3 k v$$ Given that the speed of the hailstone at time \(t = 0\) is \(u\),
3. find an expression for \(v\) in terms of \(t\).
4. Hence show that the speed of the hailstone approaches the limiting value \(\frac { g } { 3 k }\).