A ball is projected forward from a fixed point, P, on a horizontal surface with an initial speed $u ext{ m s}^{-1}$, at an acute angle $θ$ above the horizontal - AQA - A-Level Maths Pure - Question 17 - 2020 - Paper 2

Next, we will model the horizontal displacement. The horizontal distance $x$ traveled by the ball can be expressed as: $x = u t ext{ cos } θ$ Substituting the expression for $t$ gives: $x = u \left(\frac{2u \text{ sin } θ}{g}\right) ext{ cos } θ$ This simplifies to: $x = \frac{2u^{2} \text{ sin } θ ext{ cos } θ}{g}$

Since the ball must land at least $d$ metres away, we have: $x \geq d$ Thus, we can set up the inequality: $\frac{2u^{2} \text{ sin } θ ext{ cos } θ}{g} \geq d$

Using the identity for sine, $ext{sin } 2θ = 2 \text{ sin } θ \text{ cos } θ$, we can rewrite the inequality: $\frac{u^{2} \text{ sin } 2θ}{g} \geq d$ This leads us to: $\text{sin } 2θ \geq \frac{dg}{u^{2}}$