Reachable area of cannonball given fixed initial speed [closed]

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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closed as too localized by Waffle's Crazy Peanut, twistor59, akhmeteli, Manishearth♦May 5 '13 at 14:21

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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? – ABC 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

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

$$y=x.\tan\theta-\dfrac{gx^2}{2v_0^2}-\dfrac{gx^2\tan^2\theta}{2v_0^2}$$

$$A\tan^2\theta+B\tan\theta+C=0$$

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.

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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. – ABC May 3 '13 at 16:48