 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2017-resit/8/c/solution

Height decreases at rate

\dot{h} = - v

m = m_{0} - α t

\int_{0}^{t} lo g \frac{m _{0}}{m _{0} - α s} d s = t - \frac{m}{α} lo g \frac{m _{0}}{m} .

Consequently

h (t) = h_{0} - (v_{0} - u) t - \frac{1}{2} g t^{2} - \frac{u ( m _{0} - α t )}{α} lo g \frac{m _{0}}{m _{0} - α t} .

Differentiating this expression recovers

\dot{h} = - v

and substituting

t = 0

gives

h (0) = h_{0}

, which check both the sign and the integration constant.

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