Pre-U Pre-U 9794/2 Specimen — Question 13 11 marks

Exam BoardPre-U
ModulePre-U 9794/2 (Pre-U Mathematics Paper 2)
SessionSpecimen
Marks11
TopicProjectiles
TypeProjectile passing through given point
DifficultyStandard +0.3 This is a standard projectile motion problem requiring application of SUVAT equations and trajectory formulas. Part (i) uses the condition of horizontal impact to find velocity components through routine calculations. Part (ii) involves solving a quadratic equation for angle of projection. While requiring careful algebraic manipulation and understanding of projectile motion principles, both parts follow well-established methods without requiring novel insight or particularly complex reasoning.
Spec3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model

A gunner fires one shell from each of two guns on a stationary ship towards a vertical cliff \(AB\) of height \(100\) m whose foot \(A\) is at a horizontal distance \(600\) m from the point of projection.
  1. Given that the shell from the first gun hits the cliff, travelling horizontally, at a point \(45\) m above \(A\), determine the initial velocity of the shell. Express your answer in the form \(a\mathbf{i} + b\mathbf{j}\), where \(a\) and \(b\) are integers. [6]
  2. The shell from the second gun hits the cliff at its top point \(B\). Given that the initial speed of the shell is \(300\) m s\(^{-1}\), determine the possible angles of projection. [5]

The usual formulae to be used as appropriate: \(v_x = V\cos\alpha\), \(v_y = V\sin\alpha - gt\), \(x = (V\cos\alpha)t\), \(y = (V\sin\alpha)t - \frac{1}{2}gt^2\)
AnswerMarks Guidance
(i) At impact the vertical velocity is zeroM1 Obtain \(t = \frac{V\sin\alpha}{g}\)
(ii) At \(B\): \(600 = (300\cos\theta)t\) and \(100 = (300\sin\theta)t - \frac{1}{2}gt^2\)M1 Obtain \(100 = 600\tan\theta - 5 \times 4 \times \sec^2\theta\)
The usual formulae to be used as appropriate: $v_x = V\cos\alpha$, $v_y = V\sin\alpha - gt$, $x = (V\cos\alpha)t$, $y = (V\sin\alpha)t - \frac{1}{2}gt^2$

(i) At impact the vertical velocity is zero | M1 | Obtain $t = \frac{V\sin\alpha}{g}$ | A1 | $600 = (V\cos\alpha) \frac{V\sin\alpha}{g}$ | A1 | $45 = V\sin\alpha \times \frac{V\sin\alpha}{g} - \frac{1}{2}g\left(\frac{V\sin\alpha}{g}\right)^2$ | A1 | Obtain $V\cos\alpha = 200, V\sin\alpha = 30$ | A1 | Hence $V = 200\mathbf{i} + 30\mathbf{j}$ | A1 | 6 marks

(ii) At $B$: $600 = (300\cos\theta)t$ and $100 = (300\sin\theta)t - \frac{1}{2}gt^2$ | M1 | Obtain $100 = 600\tan\theta - 5 \times 4 \times \sec^2\theta$ | A1 | Use of $\sec^2\theta = 1 + \tan^2\theta$ | M1 | $\tan\theta = \frac{30 \pm \sqrt{876}}{2} = 29.799$ or $0.201$ | A1 | $\theta = 11.4°$ or $88.1°$ | A1 | 5 marks
A gunner fires one shell from each of two guns on a stationary ship towards a vertical cliff $AB$ of height $100$ m whose foot $A$ is at a horizontal distance $600$ m from the point of projection.

\begin{enumerate}[label=(\roman*)]
\item Given that the shell from the first gun hits the cliff, travelling horizontally, at a point $45$ m above $A$, determine the initial velocity of the shell. Express your answer in the form $a\mathbf{i} + b\mathbf{j}$, where $a$ and $b$ are integers. [6]

\item The shell from the second gun hits the cliff at its top point $B$. Given that the initial speed of the shell is $300$ m s$^{-1}$, determine the possible angles of projection. [5]
\end{enumerate}

\hfill \mbox{\textit{Pre-U Pre-U 9794/2  Q13 [11]}}