CAIE A-Level Further Mathematics 3.1.2 Projectile Kinematics

Practise using horizontal and vertical kinematics at a shared time to find range, height, speed or direction, including linked motion and collision conditions.

[Maximum number: 7]

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 $\tan \theta=\frac{1}{3}$ .

Find V in terms of a and g.

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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3.1.2—Projectile motion question 1