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ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2015/1/solution

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Past exam of the physics 1 paper of the University of Oxford / 2015 / 1 / Solution by

Codex  0 2026-10-25

Take the launch point as

(x, y) = (0, 0)

and ignore air drag. In projectile motion with constant gravitational acceleration

g

, a launch at angle

θ

above the horizontal gives

x = v_{0} t cos θ, y = v_{0} t sin θ - \frac{1}{2} g t^{2} .

At the wall

x = l

, the height is

y (l) = l tan θ - \frac{g l ^{2}}{2 v _{0}^{2}} sec^{2} θ = lq - \frac{g l ^{2}}{2 v _{0}^{2}} (1 + q^{2}), q = tan θ .

This is a concave quadratic polynomial in

q

, so its unique maximum satisfies

l - g l^{2} q / v_{0}^{2} = 0

. Hence

θ_{m a x} = arctan \frac{v _{0}^{2}}{g l}, y_{m a x} (l) = \frac{v _{0}^{2}}{2 g} - \frac{g l ^{2}}{2 v _{0}^{2}} .

For

l = v_{0}^{2} / g

, the best trajectory has

θ = 4 5^{\circ}

and reaches the wall at launch height. For

l > v_{0}^{2} / g

, the mathematical maximum height is negative. If the ball must stay above level ground, it lands before reaching the wall:

v_{0}^{2} / g

is the maximum same-level projectile range. If there is no intervening ground and the wall extends below the launch height, the same optimum still applies, but its impact point is below that height.

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