A ball is projected forward from a fixed point, P, on a horizontal surface with an initial speed $u \text{ ms}^{-1}$, at an acute angle $\theta$ above the horizontal - AQA - A-Level Maths: Mechanics - Question 17 - 2020 - Paper 2

In the horizontal direction, the displacement is given by: $x = u \cos \theta \cdot t$ Substituting the expression for $t$ from the previous step, we have: $x = u \cos \theta \cdot \frac{2u \sin \theta}{g} = \frac{2u^2 \sin \theta \cos \theta}{g}$

The inequality we need to satisfy for displacement is: $x \geq d$ Substituting in our expression for $x$, we get: $\frac{2u^2 \sin \theta \cos \theta}{g} \geq d$ This simplifies to: $2u^2 \sin \theta \cos \theta \geq dg$

Applying the double angle identity, we know: $\sin 2\theta = 2 \sin \theta \cos \theta$ Thus, we can rewrite our inequality as: $u^2 \sin 2\theta \geq dg$ Rearranging gives: $\sin 2\theta \geq \frac{dg}{u^2}$, which completes the proof.