A particle Q of mass m falls under gravity in a medium, whose resistance to motion is of magnitude mkv^2, where v is the speed of Q and k is a positive constant. Given that the maximum speed of Q is U, show that U=√(g/k) (g is the acceleration due to gravity).
Given that Q is projected vertically upwards with speed V(>U), show that the speed of Q is equal to U when the height of Q above the point of projection is U^2/2g ln[1/2 (1+V^2/U^2 )].
Find in terms of U, V and g, the time taken for the speed of Q to decrease from V to U.