CAIE A-Level Further Mathematics 3.1 Motion of a Projectile

Use equation of trajectory: $y=x \tan \alpha-\frac{g x^{2}}{2 u^{2}} \sec ^{2} \alpha$ :
$$\frac{3}{8} a=3 a \times \frac{1}{3}-g \times \frac{(3 a)^{2}}{2 u^{2}}\left(1+\frac{1}{9}\right)$$

M1

$\frac{5}{8} a=\frac{5 g a}{u^{2}}, \quad u^{2}=8 g a$

A1

At least one step of working.
AG

2

For P, time of flight $T=\frac{2 u \sin \theta}{g}$

B1

For P, range $=T u \cos \theta$

B1

For Q, time of flight $=\frac{1}{2} T$, so range $=\frac{1}{2} T V \cos \alpha$

Equate: $V \cos \alpha=2 u \cos \theta(1)$

B1

For Q vertically: $0=V \sin \alpha \times \frac{1}{2} T-\frac{\frac{1}{2} g T^{2}}{4}$, so $T=\frac{4 V \sin \alpha}{g}$

M1

Equate with result for P:
$$\frac{4 V \sin \alpha}{g}=\frac{2 u \sin \theta}{g} \text { so } V \sin \alpha=\frac{1}{2} u \sin \theta \text { (2) }$$

A1

From (1) and (2): $V^{2}=u^{2}\left(4(\cos \theta)^{2}+\frac{1}{4}(\sin \theta)^{2}\right)$

M1

$$\begin{aligned} V^{2}=u^{2} \times\left(4 \times \frac{9}{10}+\frac{1}{4} \times \frac{1}{10}\right)=\frac{145}{40} u^{2} V=\sqrt{29 a g} \end{aligned}$$

A1

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