Two particles, P and Q, are projected at the same time from a fixed point X, on the ground, so that they travel in the same vertical plane - AQA - A-Level Maths Pure - Question 18 - 2021 - Paper 2

To solve this, we start by considering the vertical motion of both particles.

1. The time of flight for particle P can be expressed as: $t_P = \frac{u \sin \theta}{g}$ where $g$ is the acceleration due to gravity.

2. For particle Q, the time of flight is: $t_Q = \frac{2u \sin 2\theta}{g}$

3. Since both particles land at the same point Y, their times of flight are equal: $t_P = t_Q\Rightarrow \frac{u \sin \theta}{g} = \frac{2u \sin 2\theta}{g}$

4. Removing the common terms gives us: $\sin \theta = 2 \sin 2\theta$

5. Using the double angle identity for sine, $\sin 2\theta = 2\sin \theta \cos \theta$, we can substitute: $\sin \theta = 2 \cdot 2\sin \theta \cos \theta$ This simplifies to: $1 = 4 \cos \theta$

6. Therefore, we have: $\cos \theta = \frac{1}{4}$

7. Using the identity for cosine, $\cos 2\theta = 2\cos^2 \theta - 1$, we substitute: $\cos 2\theta = 2\left(\frac{1}{4}\right)^2 - 1 = \frac{1}{8}$