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 Mechanics - Question 18 - 2021 - Paper 2

To derive the relationship between $heta$ and $2 heta$, we will start from the equations of motion for both particles.

1. For particle P:

   • The vertical motion can be described using the following equation:

   t_{p} = rac{u ext{sin} heta}{g}

   And since it takes the same time to reach the ground:

   t_p = rac{2u ext{sin} 2 heta}{g}

   We can equate the two results:

   rac{u ext{sin} heta}{g} = rac{2u ext{sin} 2 heta}{g}

   Dividing both sides by $u$ and $g$ gives:

   $ext{sin} heta = 2 ext{sin} 2 heta$

   2. We use the double angle formula for sine:

   $ext{sin} 2 heta = 2 ext{sin} heta ext{cos} heta$

   Therefore:

   $ext{sin} heta = 2(2 ext{sin} heta ext{cos} heta)$

   Simplifying gives:

   $1 = 4 ext{cos} heta$

   Thus, we find:

   ext{cos} heta = rac{1}{4}

   3. Using the identity $ext{cos} 2 heta = 2 ext{cos}^2 heta - 1$:

   Substituting ext{cos} heta = rac{1}{4}:

   ext{cos} 2 heta = 2igg(rac{1}{4}igg)^2 - 1 = 2igg(rac{1}{16}igg) - 1 = rac{1}{8}.