Further Math A2s 3.1.2 Projectile kinematics

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A particle P is projected with speed $u \mathrm{~ms}^{-1}$ at an angle $\theta$ above the horizontal from a point O on a horizontal plane and moves freely under gravity. During its flight P passes through the point which is a horizontal distance 3 a from O and a vertical distance $\frac{3}{8} a$ above the horizontal plane. It is given that $\tan \theta=\frac{1}{3}$ .

(a)

Show that $u^{2}=8 a g$ .
A particle Q is projected with speed $V \mathrm{~ms}^{-1}$ at an angle $\alpha$ above the horizontal from O at the instant when P is at its highest point. Particles P and Q both land at the same point on the horizontal plane at the same time.

[ 2 ]

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

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(b)

Find V in terms of a and g.

[ 7 ]

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): $V^{2}=u^{2}\left(4(\cos \theta)^{2}+\frac{1}{4}(\sin \theta)^{2}\right)$ $\begin{aligned} V^{2}=u^{2} \times\left(4 \times \frac{9}{10}+\frac{1}{4} \times \frac{1}{10}\right)=\frac{145}{40} u^{2} V=\sqrt{29 a g} \end{aligned}$
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CAIE A-Level Further Math A2 3.1.2 Projectile Motion Questions

Exam points

CAIE A-Level Further Math A2 3.1.2 Projectile Motion Questions question 1

Question (a)

Question (b)