| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Topic | Projectiles |
| Type | Maximum range or optimal angle |
| Difficulty | Challenging +1.8 This is a challenging projectile motion problem requiring rearrangement into quadratic form, discriminant analysis to find the envelope of trajectories, then optimization with a shifted origin. While the individual techniques (quadratic discriminant, completing the square, differentiation) are standard A-level, the conceptual leap to the envelope of trajectories and the multi-step nature with non-trivial algebra places it significantly above average difficulty but below the most demanding proof-based questions. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02f Solve quadratic equations: including in a function of unknown3.02i Projectile motion: constant acceleration model |
It is given that the trajectory of a projectile which is launched with speed $V$ at an angle $\alpha$ above the horizontal has equation
$$y = x\tan\alpha - \frac{gx^2}{2V^2}(1 + \tan^2\alpha),$$
where the point of projection is the origin, and the $x$- and $y$-axes are horizontal and vertically upwards respectively.
\begin{enumerate}[label=(\roman*)]
\item Express the above equation as a quadratic equation in $\tan\alpha$ and deduce that the boundary of all accessible points for this projectile has equation
$$y = \frac{1}{2gV^2}(V^4 - g^2x^2).$$ [4]
\item A stone is thrown with speed $\sqrt{gh}$ from the top of a vertical tower, of height $h$, which stands on horizontal ground. Find
\begin{enumerate}[label=(\alph*)]
\item the maximum distance, from the foot of the tower, at which the stone can land, [3]
\item the angle of elevation at which the stone must be thrown to achieve this maximum distance. [3]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2014 Q11 [10]}}