| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Session | Specimen |
| Marks | 13 |
| Topic | Projectiles |
| Type | Deriving trajectory equation |
| Difficulty | Standard +0.8 This is a multi-part projectile motion question requiring derivation of standard equations, then applying calculus and geometry to find relationships between angles. Part (i) is routine, but parts (ii)(b) and (ii)(c) require insight connecting the angle of the position vector to velocity components and maximum height conditions, with algebraic manipulation across multiple steps. More demanding than typical A-level mechanics but not exceptionally difficult for Further Maths students. |
| Spec | 1.05q Trig in context: vectors, kinematics, forces3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
**(i)** Applying $s = ut - \frac{1}{2}at^2$ to the horizontal and vertical motions: M1
$x = V\cos\alpha\, t$ A1
$y = V\sin\alpha\, t - \frac{1}{2}gt^2$ A1 **[3]**
**(ii)(a)** Applying the trigonometric definition of the tangent of an angle: M1
$\tan\theta = \frac{y}{x} = \frac{V\sin\alpha\, t - \frac{1}{2}gt^2}{V\cos\alpha\, t}$ A1
$= \tan\alpha - \frac{g}{2V\cos\alpha}\,t$ **AG** **[2]**
**(ii)(b)** The maximum height is attained when $\frac{dy}{dt} = 0$: M1
$\Rightarrow t = \frac{V\sin\alpha}{g}$ A1
$\tan\beta = \tan\alpha - \frac{g}{2V\cos\alpha} \times \frac{V\sin\alpha}{g}$ A1
$\Rightarrow \tan\beta = \frac{1}{2}\tan\alpha$ **AG** **[3]**
**(ii)(c)** If the particle strikes the ground at time $T$ then
$\tan\beta = \frac{H}{VT\cos\alpha}$ M1
$\Rightarrow T = \frac{H}{V\tan\beta\cos\alpha} \left(= \frac{2H}{V\sin\alpha}\right)$ A1
Now using $s = ut - \frac{1}{2}at^2$: M1
$-H = V\sin\alpha\, T - \frac{1}{2}gT^2$ A1
$-H = 2H - \frac{2gH^2}{V^2\sin^2\alpha}$ A1
$\Rightarrow H = \frac{3V^2\sin^2\alpha}{2g}$ **AG** **[5]**14 A particle $P$ is projected from the point $O$, at the top of a vertical wall of height $H$ above a horizontal plane, with initial speed $V$ at an angle $\alpha$ above the horizontal. At time $t$ the coordinates of the particle are $( x , y )$ referred to horizontal and vertical axes at $O$.
\begin{enumerate}[label=(\roman*)]
\item Express $x$ and $y$ as functions of $t$.
Let $\theta$ be the angle $O P$ makes with the horizontal at time $t$.
\item (a) Show that
$$\tan \theta = \tan \alpha - \frac { g } { 2 V \cos \alpha } t$$
(b) Show that when the particle attains its greatest height above the point of projection, where $O P$ makes an angle $\beta$ with the horizontal,
$$\tan \beta = \frac { 1 } { 2 } \tan \alpha .$$
(c) If the particle strikes the ground where $O P$ makes an angle $\beta$ below the horizontal, show that
$$H = \frac { 3 V ^ { 2 } \sin ^ { 2 } \alpha } { 2 g }$$
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 Q14 [13]}}