 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2017-resit/7/solution

Choose upward velocity positive during ascent. Gravity and linear drag both point downward, giving the linear differential equation

m \overset{v}{˙} = - m g - αv, v (t) = (v_{0} + \frac{m g}{α}) e^{- α t / m} - \frac{m g}{α} .

At the highest point

v = 0

, so

t_{m a x} = \frac{m}{α} lo g (1 + \frac{α v _{0}}{m g}) .

v_{0} / g

for

α > 0

, because

lo g (1 + x) < x

. Its zero-drag limit is

v_{0} / g

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