| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/3 (Pre-U Mathematics Paper 3) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Topic | Projectiles |
| Type | Projectile passing through given point |
| Difficulty | Standard +0.3 This is a straightforward projectile motion problem requiring standard kinematic equations. Part (i) involves substituting given values into x = ut cos θ and y = ut sin θ - ½gt² to find two equations in two unknowns, then solving simultaneously. Part (ii) is a direct application of the range formula once u and θ are known. While it requires careful algebraic manipulation, it follows a well-practiced procedure with no novel insight needed, making it slightly easier than average. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
**Question 7**
**(i)** Horiz: $18 = 2u\cos\theta$ — B1: Use of $x = ut\cos\theta$
Vert: $4 = 2u\sin\theta - 20$ — B1: Use of $y = ut\sin\theta - \frac{1}{2}gt^2$
$\therefore u\cos\theta = 9$ and $u\sin\theta = 12$ — M1: Attempt to eliminate $u$.
$\therefore \tan\theta = \frac{12}{9} = \frac{4}{3}$ — A1: A.G. Convincingly shown.
$u^2 = 9^2 + 12^2 = 225$
$\therefore u = 15\ \text{ms}^{-1}$ — M1: Eliminate or substitute for $\theta$. Allow $u$ found first then $\theta$ using $u$ provided it does not involve a circular argument. A1: c.a.o. **[6]**
**(ii)** $R = \frac{2u^2}{g}\sin\theta\cos\theta = \frac{2 \times 15^2}{10} \times \frac{4}{5} \times \frac{3}{5}$
$= 21.6\ \text{m}$
— M1: Use of formula for range, or equivalent. A1: Ft *their* $u$. **[2]**7 A particle is projected from the origin with initial speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle $\theta$ above the horizontal. After 2 seconds the particle is at a point which is 18 m horizontally from the origin and 4 m above it.\\
(i) Show that $\tan \theta = \frac { 4 } { 3 }$ and find $u$.\\
(ii) Find the horizontal range of the particle.
\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2015 Q7 [8]}}