2022TrialShort answer

Q16 (c) (i)

2 marks

Question

A projectile is launched from $O$ at $V\,\mathrm{m/s}$ so that it will hit the point $A(a,b)$ on the slope. Air resistance is neglected and the acceleration due to gravity is $g\,\mathrm{m/s^2}$ . Given that
$y=-\frac{gx^2}{2V^2}(1+\tan^2\alpha)+x\tan\alpha,$
and that $\alpha_1<\alpha_2$ , show that
$\tan\alpha_1=\frac{V^2-\sqrt{V^4-g^2a^2-2V^2gb}}{ga}$
and
$\tan\alpha_2=\frac{V^2+\sqrt{V^4-g^2a^2-2V^2gb}}{ga}.$

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Solution

Q.16(c)(i) Hits target at $(a,b)$ .

$y=-\frac{gx^2}{2V^2}(1+\tan^2\alpha)+x\tan\alpha$

$b=-\frac{ga^2}{2V^2}-\frac{ga^2}{2V^2}\tan^2\alpha+a\tan\alpha$

$0=ga^2\tan^2\alpha-2V^2a\tan\alpha+(ga^2+2V^2b)$

$\therefore\tan\alpha=\frac{2V^2a\pm\sqrt{4V^4a^2-4ga^2(ga^2+2V^2b)}}{2ga^2}$

$=\frac{2V^2a\pm\sqrt{4a^2\left[V^4-g^2a^2-2gV^2b\right]}}{2ga^2}$

$=\frac{2V^2a\pm2a\sqrt{V^4-g^2a^2-2gV^2b}}{2ga^2}$

$\tan\alpha=\frac{V^2\pm\sqrt{V^4-g^2a^2-2gV^2b}}{ga}$

As $\alpha_2>\alpha_1$ ,
$\tan\alpha_2=\frac{V^2+\sqrt{V^4-g^2a^2-2gV^2b}}{ga}$
$\&\ \tan\alpha_1=\frac{V^2-\sqrt{V^4-g^2a^2-2gV^2b}}{ga}.$

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Mechanics

Projectiles and resisted motionUse parametric and Cartesian equations to solve projectile problems with and without resistance.

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