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The trajectory of a cannonball fired from the origin with initial speed $v_0$ at an angle $\theta$ is given by $$y = x\tan\theta - \frac{g}{2v_0^2 \cos^2\theta}x^2.$$ For fixed $v_0$, at what angle $\theta$ should the canon ball be fired so that it will hit a target located at $(x_0,y_0)$? What is the area of the region in which the target can be located such that it is possible to hit the target?

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I'm a little confused: what do you mean by "field"? –  David Z May 3 '13 at 3:11
Same here.If you want to say that max. area that the ball can cover at different angles? –  Mr.ØØ7 May 3 '13 at 3:12
@DavidZaslavsky Contour might be a better word to use, for theta from 0 to 180 degree, the furthest point the cannonball can hit can be connected as a contour line. –  Arch1tect May 3 '13 at 3:14
@007 I think you get what I mean max area is right:) –  Arch1tect May 3 '13 at 3:15
Consider using $\frac{1}{\cos^2\theta}=1+\tan^2\theta$ to obtain a quadratic equation in $\tan\theta$, with coefficients dependent on $x,y,v_0$. –  leongz May 3 '13 at 6:13
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closed as too localized by Waffle's Crazy Peanut, twistor59, akhmeteli, Manishearth May 5 '13 at 14:21

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1 Answer

up vote 3 down vote accepted

$$y = x\tan\theta - \frac{g}{2v_0^2 \cos^2\theta}x^2.$$



As $\tan\theta\in\Bbb R$ , so $B^2-4AC\ge0$ must hold, for the above equation to have real roots for $\tan\theta$.

Use that and you'll get $$y\le \dfrac{v^2}{2g}-\dfrac{g}{2v^2}.x^2 $$

That defines a area under the parabola , within which any target can be hit.

enter image description here

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OMG, exactly what I want...You mind reader:) –  Arch1tect May 3 '13 at 16:42
Is it B^2 - AC or B^2 - 4AC ? –  Arch1tect May 3 '13 at 16:47
@Arch1tect Hmm... Yes $4AC$ the $ sign ate up the 4. Thanx. –  Mr.ØØ7 May 3 '13 at 16:48
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