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The range of a particle projected at an angle $\alpha$ from a horizontal plane, neglecting air resistance is $\dfrac{v_0^2sin(2\alpha)}{g}$. The particle falls short of the target by a distance of $a$. When the particle is projected at an angle $\beta$, the falls beyond the target by a distance of $b$. If $\theta$ is the angle of projection required to hit the target directly, show that

$(a+b)sin(2\theta)$ = $asin(2\beta) + bsin(2\alpha)$

I'm not really sure how to go about this question. Any help would be much appreciated.

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1 Answer 1

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I am surprised there is no answer to this question yet.

To solve the question, we will first of all ignore $\frac{v_0^2}{g}$, since the answer won't depend on these. The only effect these terms would have, would be to obscure what's actually going on. If you want to include them, feel free to, as long as you ensure they cancel out at the end.

We will now say the projectiles are launched from $x=0$ and that the target is as $x=t$. The angle needed to hit the target is $\theta$. Since we know what happens for two angles, we now get the following equations: $$t-a = \sin(2\alpha),\\ t+b = \sin(2\beta),\\ t = \sin(2\theta).$$

Realising that $$\frac{a(t+b)+b(t-a)}{a+b} = \frac{at+ab+bt-ab}{a+b} = \frac{(a+b)t}{a+b}=t,$$ you can use these three equations to obtain $$(a+b)\sin(2\theta) = (a+b)t = a(t+b)+b(t-a) = a\sin(2\beta) + b\sin(2\alpha).$$

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