6. A particle of mass \(m\) is projected vertically upwards in a resisting medium. As the particle moves upwards, the speed \(v\) of the particle is given by $$v ^ { 2 } = k g \left( 5 \mathrm { e } ^ { - \frac { x } { 2 k } } - 4 \right)$$ where \(x\) is the distance of the particle above the point of projection and \(k\) is a positive constant.

1. Show that the magnitude of the resistance to the motion of the particle is \(\frac { m v ^ { 2 } } { 4 k }\).
   (4)
2. Find, in terms of \(k\), the greatest height reached by the particle above the point of projection.
3. Show that the time taken by the particle to reach its greatest height above the point of projection is \(\sqrt { \frac { 4 k } { g } } \arctan \left( \frac { 1 } { 2 } \right)\)