At time \(t = 0\), a small stone \(P\) of mass \(m\) is released from rest and falls vertically through the air. At time \(t\), the speed of \(P\) is \(v\) and the resistance to the motion of \(P\) from the air is modelled as a force of magnitude \(k v ^ { 2 }\), where \(k\) is a constant.
Show that \(t = \frac { V } { 2 g } \ln \left( \frac { V + v } { V - v } \right)\) where \(V ^ { 2 } = \frac { m g } { k }\)
Give an interpretation of the value of \(V\), justifying your answer.
At time \(t , P\) has fallen a distance \(s\).
Show that \(s = \frac { V ^ { 2 } } { 2 g } \ln \left( \frac { V ^ { 2 } } { V ^ { 2 } - v ^ { 2 } } \right)\)