| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Finding angle given constraints |
| Difficulty | Standard +0.3 This is a standard M2 projectile question requiring derivation of the trajectory equation followed by solving a quadratic in tan θ. Part (a) is routine application of SUVAT equations, and part (b) involves straightforward algebraic substitution and solving a quadratic—slightly easier than average due to clear guidance and standard techniques. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(x = (u \cos \theta)t\), \(y = h + (u \sin \theta)t - \frac{1}{2}gt^2\) | B1 M1 A1 | |
| \(y = h + x \tan \theta - \frac{g}{2u^2\cos^2 \theta}x^2\) | M1 A1 | |
| \(0 = h + d \tan \theta - \frac{gd^2}{2u^2\cos^2 \theta}\) | M1 | |
| \(\frac{gd^2}{2u^2} \sec^2 \theta - d \tan \theta - h = 0\) | A1 | |
| (b) Let \(\tan \theta = T\) | M1 A1 | |
| Subst. given values: \(4.9(1 + T^2) - 147T - 7 = 0\) | ||
| \(7T^2 - 20T - 3 = 0\) | ||
| \((7T + 1)(T - 3) = 0\) | M1 A1 | |
| \(T = 3\) | ||
| \(\theta = 71.7°\) | A1 A1 | 12 marks |
(a) $x = (u \cos \theta)t$, $y = h + (u \sin \theta)t - \frac{1}{2}gt^2$ | B1 M1 A1 |
$y = h + x \tan \theta - \frac{g}{2u^2\cos^2 \theta}x^2$ | M1 A1 |
$0 = h + d \tan \theta - \frac{gd^2}{2u^2\cos^2 \theta}$ | M1 |
$\frac{gd^2}{2u^2} \sec^2 \theta - d \tan \theta - h = 0$ | A1 |
(b) Let $\tan \theta = T$ | M1 A1 |
Subst. given values: $4.9(1 + T^2) - 147T - 7 = 0$ | |
$7T^2 - 20T - 3 = 0$ | |
$(7T + 1)(T - 3) = 0$ | M1 A1 |
$T = 3$ | |
$\theta = 71.7°$ | A1 A1 | 12 marks\begin{enumerate}
\item A ball is hit with initial speed $u \mathrm {~ms} ^ { - 1 }$, at an angle $\theta$ above the horizontal, from a point at a height of $h \mathrm {~m}$ above horizontal ground. The ball, which is modelled as a particle moving freely under gravity, hits the ground at a horizontal distance $d \mathrm {~m}$ from the point of projection.\\
(a) Prove that $\frac { g d ^ { 2 } } { 2 u ^ { 2 } } \sec ^ { 2 } \theta - d \tan \theta - h = 0$.
\end{enumerate}
Given further that $u = 14 , h = 7$ and $d = 14$, and assuming the result $\sec ^ { 2 } \theta = 1 + \tan ^ { 2 } \theta$,\\
(b) find the value of $\theta$.\\
\hfill \mbox{\textit{Edexcel M2 Q6 [12]}}