Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 4 / 4A / Solution

Measure displacement downward from the release point and take downward speed

v \geq 0

. Newton's second law with quadratic drag gives

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

V = \frac{m g}{γ}, k = \frac{g γ}{m},

separation of variables gives the terminal velocity solution

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}) .

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