CAIE A-Level Further Math AS 3.1 Motion of a Projectile Questions

Use equation of trajectory: $y=x \tan \alpha-\frac{g x^{2}}{2 u^{2}} \sec ^{2} \alpha$ :
$$\frac{3}{8} a=3 a \times \frac{1}{3}-g \times \frac{(3 a)^{2}}{2 u^{2}}\left(1+\frac{1}{9}\right)$$$\frac{5}{8} a=\frac{5 g a}{u^{2}}, \quad u^{2}=8 g a$
At least one step of working.
AG

2

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

Equate: $V \cos \alpha=2 u \cos \theta(1)$
For Q vertically: $0=V \sin \alpha \times \frac{1}{2} T-\frac{\frac{1}{2} g T^{2}}{4}$, so $T=\frac{4 V \sin \alpha}{g}$
Equate with result for P:
$$\frac{4 V \sin \alpha}{g}=\frac{2 u \sin \theta}{g} \text { so } V \sin \alpha=\frac{1}{2} u \sin \theta \text { (2) }$$
From (1) and (2): $V^{2}=u^{2}\left(4(\cos \theta)^{2}+\frac{1}{4}(\sin \theta)^{2}\right)$$$\begin{aligned} V^{2}=u^{2} \times\left(4 \times \frac{9}{10}+\frac{1}{4} \times \frac{1}{10}\right)=\frac{145}{40} u^{2} V=\sqrt{29 a g} \end{aligned}$$
7

$\rightarrow x=u \cos \alpha t$$$\uparrow y=u \sin \alpha t-\frac{1}{2} g t^{2}$$
Both correct.

Eliminate t: $y=u \sin \alpha \times \frac{x}{\mathrm{u} \cos \alpha}-\frac{1}{2} g\left(\frac{x}{\mathrm{u} \cos \alpha}\right)^{2}$
Eliminate.

$y=x \tan \alpha-\frac{g x^{2}}{2 u^{2}} \sec ^{2} \alpha \quad \mathrm{AG}$
3

$\frac{4}{5} h=32 \tan \alpha-\frac{g 32^{2}}{2 \times 40^{2}} \sec ^{2} \alpha \quad\left[=32 \tan \alpha-\frac{16}{5} \sec ^{2} \alpha\right]$$h=32 \tan \alpha-\frac{g 32^{2}}{2(40 \sqrt{2})^{2}} \sec ^{2} \alpha\left[=32 \tan \alpha-\frac{8}{5} \sec ^{2} \alpha\right]$$$\begin{aligned} 32 \tan \alpha-\frac{8}{5} \sec ^{2} \alpha=\frac{5}{4}\left(32 \tan \alpha-\frac{16}{5} \sec ^{2} \alpha\right) 32 t-\frac{8}{5}\left(1+t^{2}\right)=40 t-4\left(1+t^{2}\right) 3 t^{2}-10 t+3=0 \end{aligned}$$
Equate expressions for h and obtain a 3 -term quadratic in
$\tan \alpha$.

$t=3, \quad \frac{1}{3}$
Both correct
For using their value of t to work out one value of h.

$$\begin{aligned} h=80 h=\frac{80}{9} \end{aligned}$$
Both correct

6

$H=80 \sin ^{2} \theta$ or $\frac{800 \sin ^{2} \theta}{g}$$T=4 \sin \theta$ or $\frac{40 \sin \theta}{g}$
2

Between t=T and $t=3 \quad \uparrow \quad \frac{1}{4} H=\frac{1}{2} \times 10 \times(3-T)^{2}$
No extra terms.

Use results from part (a) $\frac{1}{4} 80 \sin ^{2} \theta=5(3-4 \sin \theta)^{2}$$$4 \sin ^{2} \theta-8 \sin \theta+3=0$$
Substitute their expressions for H and T from
part (a) and obtain a quadratic equation in $\sin \theta$
with no more than three terms.

$\sin \theta=\frac{1}{2}, \theta=30^{\circ}$
Single answer. NFWW.

Alternative method for question 7 part (b)

$\frac{3}{4} H=y(3)=40 \times 3 \sin \theta-\frac{1}{2} \times 10 \times 3^{2}$$120 \sin \theta-45$

Use results from (a): $\frac{3}{4} 80 \sin ^{2} \theta=120 \sin \theta-45$$$4 \sin ^{2} \theta-8 \sin \theta+3=0$$
Substitute their expressions for H and T from
part (a) and obtain a quadratic equation in $\sin \theta$
with no more than three terms.

$\sin \theta=\frac{1}{2}, \theta=30^{\circ}$
Single answer. NFWW.

4

When t=3 speeds $\rightarrow 40 \cos \theta$ and $\uparrow 40 \sin \theta-10 \times 3$
Square and add to find square of speed: $v^{2}=(20 \sqrt{3})^{2}+(-10)^{2}$
Must be numerical.

$v^{2}=1300, \quad v=10 \sqrt{13} \quad[=36.1]$
3