4 A parachutist of mass 90 kg falls vertically from rest. The forces acting on her are her weight and resistance to motion \(R \mathrm {~N}\). At time \(t \mathrm {~s}\) the velocity of the parachutist is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and the distance she has fallen is \(x \mathrm {~m}\). While the parachutist is in free-fall (i.e. before the parachute is opened), the resistance is modelled as \(R = k v ^ { 2 }\), where \(k\) is a constant. The terminal velocity of the parachutist in free-fall is \(60 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. Show that \(k = \frac { g } { 40 }\).
2. Show that \(v ^ { 2 } = 3600 \left( 1 - \mathrm { e } ^ { - \frac { g x } { 1800 } } \right)\). When she has fallen 1800 m , she opens her parachute.
3. Calculate, by integration, the work done against the resistance before she opens her parachute. Verify that this is equal to the loss in mechanical energy of the parachutist. As the parachute opens, the resistance instantly changes and is now modelled as \(R = 90 v\).
4. Calculate her velocity just before opening the parachute, correct to four decimal places.
5. Formulate and solve a differential equation to calculate the time it takes after opening the parachute to reduce her velocity to \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).