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## Homework Statement

A projectile is thrown from a point P. It moves in such a way that its distance from P is always increasing. Find the maximum angle above the horizontal with which the particle could have been thrown. Ignore air resistance.

## Homework Equations

[tex]y = v_{0}\sin{(\alpha_{0})}t - \frac{1}{2}gt^2[/tex]

[tex]x = v_{0}\cos{(\alpha_{0})}t[/tex]

## The Attempt at a Solution

I've looked around on here and found two threads documenting the same problem, however one was poorly explained (no algebra), and the other was misinterpreted as a range problem.

Distance from origin:

[tex]d(t, \alpha_{0}) = \sqrt{x^2+y^2}

= t\sqrt{v_{0}^2-gtv_{0}\sin{(\alpha_{0})}+\frac{1}{4}g^2t^2}[/tex]

I've tried working from here on paper but really can't get anywhere. Differentiating wrt t gets me an inequality in terms of [tex]\alpha_{0}[/tex], [tex]v_{0}[/tex], [tex]g[/tex] and [tex]t[/tex].. but I don't know where to go from there anyway.

Any help would be appreciated. I have a feeling I'm looking at it the wrong way...

[tex]\frac{\partial d}{\partial t}(t, \alpha_{0}) = \frac{2(v_{0}^2-gtv_{0}\sin{(\alpha_{0})}+\frac{1}{4}g^2t^2)+t(\frac{1}{2}g^2t-gv_{0}\sin{(\alpha_{0})})}{2\sqrt{v_{0}^2-gtv_{0}\sin{(\alpha_{0})}+\frac{1}{4}g^2t^2}} = \frac{2v_{0}^2-3gtv_{0}\sin{(\alpha_{0})}+g^2t^2}{2\sqrt{v_{0}^2-gtv_{0}\sin{(\alpha_{0})}+\frac{1}{4}g^2t^2}}[/tex]

So:

[tex]2v_{0}^2-3gtv_{0}\sin{(\alpha_{0})}+g^2t^2 > 0[/tex]

[tex]\alpha_{0} < \arcsin{\left(\frac{2v_{0}}{3gt}+\frac{gt}{3v_{0}}\right)}[/tex]

Where it's known that:

[tex]v_{0}^2-gtv_{0}\sin{(\alpha_{0})}+\frac{1}{4}g^2t^2 > 0[/tex]

I honestly have no ****ing idea if I'm even close.

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