Past exam of the physics 1 paper of the University of Oxford / 2013 resit / 1 / Solution

Take the upward velocity as positive during ascent. Newton's second law with linear drag givesMultiplication by the integrating factor $e^{kt}$ and integration giveThe maximum height occurs at the first zero of this velocity, henceHere $k$ has units of inverse time. The limit $k\to 0$ gives $v_0/g$, and positive drag shortens the ascent relative to motion in a vacuum. The expression applies only up to the turning point; the drag force opposes the subsequent downward motion.