9 A particle P of mass 5 kg is released from rest at a point O and falls vertically. A resistance of magnitude \(0.05 v ^ { 2 } \mathrm {~N}\) acts vertically upwards on P , where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the velocity of P when it has fallen a distance \(x\) m.
- Show that \(\left( \frac { 100 \mathrm { v } } { 980 - \mathrm { v } ^ { 2 } } \right) \frac { \mathrm { dv } } { \mathrm { dx } } = 1\).
- Verify that \(\mathrm { v } ^ { 2 } = 980 \left( 1 - \mathrm { e } ^ { - 0.02 \mathrm { x } } \right)\).
- Determine the work done against the resistance while P is falling from O to the point where P's acceleration is \(8.36 \mathrm {~m} \mathrm {~s} ^ { - 2 }\).