4. Whilst in free-fall a parachutist falls vertically such that his velocity, \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), when he is \(x\) metres below his initial position is given by $$v ^ { 2 } = k g \left( 1 - \mathrm { e } ^ { - \frac { 2 x } { k } } \right) ,$$ where \(k\) is a constant.
Given that he experiences an acceleration of \(\mathrm { f } \mathrm { m } \mathrm { s } ^ { - 2 }\),

1. show that \(f = g \mathrm { e } ^ { - \frac { 2 x } { k } }\). After falling a large distance, his velocity is constant at \(49 \mathrm {~ms} ^ { - 1 }\).
2. Find the value of \(k\).
3. Hence, express \(f\) in the form ( \(\lambda - \mu v ^ { 2 }\) ) where \(\lambda\) and \(\mu\) are constants which you should find.
   (4 marks)