Further Math A2s 3.1.3 Projectile trajectories

[Maximum number: 9]

A particle P is projected with speed u at an angle $\alpha$ above the horizontal from a point O on a horizontal plane and moves freely under gravity. The horizontal and vertical displacements of P from O at a subsequent time t are denoted by x and y respectively.

(a)

Derive the equation of the trajectory of P in the form

y=x \tan \alpha-\frac{g x^{2}}{2 u^{2}} \sec ^{2} \alpha

During its flight, P must clear an obstacle of height $h \mathrm{~m}$ that is at a horizontal distance of 32 m from the point of projection. When $u=40 \sqrt{2} \mathrm{~ms}^{-1}, P$ just clears the obstacle. When $u=40 \mathrm{~ms}^{-1}, P$ only achieves 80% of the height required to clear the obstacle.

[ 3 ]

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

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

Find the two possible values of h.

[ 6 ]

$\frac{4}{5} h=32 \tan \alpha-\frac{g 32^{2}}{2 \times 40^{2}} \sec ^{2} \alpha \quad\left[=32 \tan \alpha-\frac{16}{5} \sec ^{2} \alpha\right]$ $h=32 \tan \alpha-\frac{g 32^{2}}{2(40 \sqrt{2})^{2}} \sec ^{2} \alpha\left[=32 \tan \alpha-\frac{8}{5} \sec ^{2} \alpha\right]$ $\begin{aligned} 32 \tan \alpha-\frac{8}{5} \sec ^{2} \alpha=\frac{5}{4}\left(32 \tan \alpha-\frac{16}{5} \sec ^{2} \alpha\right) 32 t-\frac{8}{5}\left(1+t^{2}\right)=40 t-4\left(1+t^{2}\right) 3 t^{2}-10 t+3=0 \end{aligned}$
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

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CAIE A-Level Further Math A2 3.1.3 Projectile Motion Questions

Exam points

CAIE A-Level Further Math A2 3.1.3 Projectile Motion Questions question 1

Question (a)

Question (b)