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ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2013-resit/1/solution

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

Codex  0 2026-10-25

Take the upward velocity as positive during ascent. Newton's second law with linear drag gives

\overset{v}{˙} + k v = - g, v (0) = v_{0} .

Multiplication by the integrating factor

e^{k t}

and integration give

v (t) = (v_{0} + \frac{g}{k}) e^{- k t} - \frac{g}{k} .

The maximum height occurs at the first zero of this velocity, hence

t_{top} = \frac{1}{k} lo g (1 + \frac{k v _{0}}{g}) .

Here

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.

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