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The range of a dragless projectile fired at angle $\theta$ above the horizontal with an an initial height can be written non-dimensionally as:

$$R^* = \cos \theta \left (\sin \theta + \sqrt{\sin^2 \theta + 2 / Fr}\right)$$

where $R^* = R g / v$, $Fr = v^2 / (g y_0)$, $v$ is the initial velocity, and $y_0$ is the initial height.

(Derivation is available on Wikipedia, in a different notation than mine here. I use dimensionless notation to simplify the equation.)

The firing angle which maximizes the range is also given on Wikipedia and elsewhere, but I am unable to derive it aside from a special case ($y_0 = 0$). In my notation, the optimal firing angle can be found from:

$$\sin^2 \theta_{max} = \frac{Fr}{2 (Fr + 1)}$$

Differentiating $R^*$ with respect to $\theta$ is simple enough, as is using the result above to check that the $d R / d \theta = 0$ for the optimal angle. That proves the result. But how do you derive this in the first place? My own attempts at solving $d R / d \theta = 0$ for $\theta$ get lost in a mess of trigonometric functions, and I'm interested in seeing how one cuts through that mess to get the result.

Ben Trettel
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    As I remember from years ago, It'll take you a page or two, but go through it neatly and it works out. It might be cleaner if you let $2/Fr = a$, and only substitute back at the very end. The whole exercise is pretty masochistic, though. – stochasticboy321 Oct 16 '15 at 01:36

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The trajectory has equation $$y=x \tan \theta- \frac {g x^2}{2v^2} \sec^2 \theta$$ Then the equation $y=-y_0$, with $y_0>0$, defines (implicitly) the range of interest $x=R$ as a function of $\theta$.

Now differentiate $y=-y_0$ with respect to $\theta$ obtaining $$\frac {dR}{d\theta}\tan \theta+R \sec^2 \theta- \frac {gR}{v^2} \frac {dR}{d\theta} \sec^2 \theta- \frac {gR^2}{v^2} \sec^2 \theta \tan \theta=0$$ Then put $$\frac {dR}{d\theta}=0$$ obtaining $$R=\frac{v^2}g \cot \theta$$ Substitute the latter in $y=-y_0$ to get $$\sin^2 \theta= \frac {v^2}{2v^2+2gy_0}$$

Tony Piccolo
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