A particle is projected upwards in a medium whose resistance is gv/V, where v is the velocity. If V is large compared to U, which is the velocity of projection, show that the fraction of the value of the vertical height h reached by the particle is described as
2U/3V
and the fraction of the value of the ascent time is
U/2V
and the fraction of the value of the descent time is
U/6V
when there is no resistance.
V can be taken as the terminal velocity of the particle, although the question does not state this. The particle could reach the terminal velocity on its way upwards, before its return journey.
Prove that the particle returns to the point of projection with the velocity
U(1−2U/3V)
The first 2 parts done by solving dv/dt = -g - gv/V and using Taylor expansions for the log parts of the solutions and simplifying.
Getting U/6V should be obtainable by solving dv/dt = g - gv/V or is there another way?