 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-4/4a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Measure displacement downward from the release point and take downward speed

v \geq 0

m \frac{d v}{d t} = m g - γ v^{2}, v (0) = 0.

V = \frac{m g}{γ}, k = \frac{g γ}{m},

v (t) = V tanh (k t) .

The distance fallen by time

t

is therefore

y (t) = \int_{0}^{t} v (s) d s = \frac{m}{γ} lo g cosh (k t) .

Setting

y (t) = h

and solving for

t

yields

t = \frac{m}{g γ} arcosh (e^{γh / m}) .

 New to topics? Read the docs here!