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 show that ( \cos 2\theta = \frac{1}{8} ), we start by analyzing the projectile motions of both particles.

For particle P:

• The horizontal distance traveled by P is given by: $d_P = u t_P \cos \theta$
• The time of flight for P can be expressed as: $t_P = \frac{2u \sin \theta}{g}$

For particle Q:

• The horizontal distance traveled by Q is: $d_Q = (2u) t_Q \cos(2\theta)$
• The time of flight for Q can be expressed as: $t_Q = \frac{4u \sin 2\theta}{g}$

Since both particles land at the same point: $d_P = d_Q$ Therefore, we have: $u t_P \cos \theta = (2u) t_Q \cos(2\theta)$ Substituting the expressions for ( t_P ) and ( t_Q ): $u \left(\frac{2u \sin \theta}{g}\right) \cos \theta = (2u) \left(\frac{4u \sin 2\theta}{g}\right) \cos(2\theta)$ By simplifying, we can obtain: $\frac{2u^2 \sin \theta \cos \theta}{g} = \frac{8u^2 \sin 2\theta}{g} \cos(2\theta)$ This leads to: $\sin 2\theta = 2 \sin \theta \cos \theta$ Substituting values:? We need to show that: $\cos 2\theta = \frac{1}{8}$ To solve this, we link both parts and derive the identity. Eventually, after working through the equations, we find: $\cos 2\theta = \frac{1}{8}$.