Particle $P$ is fired with speed $V\,\text{m s}^{-1}$ at $60^{\circ}$ above the horizontal from point $O$. Exactly $1\,\text{s}$ later, particle $Q$ is launched from $O$ with the same initial speed at $45^{\circ}$ above the horizontal. The particles meet when $Q$ has been moving for $t\,\text{s}$.
(i)[3]
Show that $t = 2.414$, correct to $3$ decimal places.
(ii)[4]
Find $V$.
(iii)[4]
Because the collision takes place after $P$ has gone past the top of its path, calculate the vertical distance of $P$ below its highest point at the instant $P$ and $Q$ collide.
Worked solution & mark scheme
This 11-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Sets horizontal distances equal: $V\cos45\,t = V\cos60(t+1)$” …