Projectile Motion and maximum angle

[blindside]

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

y = v_{0}\sin{(\alpha_{0})}t - \frac{1}{2}gt^2
x = v_{0}\cos{(\alpha_{0})}t

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:

d(t, \alpha_{0}) = \sqrt{x^2+y^2}<br /> = t\sqrt{v_{0}^2-gtv_{0}\sin{(\alpha_{0})}+\frac{1}{4}g^2t^2}

I've tried working from here on paper but really can't get anywhere. Differentiating wrt t gets me an inequality in terms of \alpha_{0}, v_{0}, g and t.. 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...

\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}}

So:

2v_{0}^2-3gtv_{0}\sin{(\alpha_{0})}+g^2t^2 &gt; 0
\alpha_{0} &lt; \arcsin{\left(\frac{2v_{0}}{3gt}+\frac{gt}{3v_{0}}\right)}

Where it's known that:

v_{0}^2-gtv_{0}\sin{(\alpha_{0})}+\frac{1}{4}g^2t^2 &gt; 0

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

 

[psi^]

d^2 = x^2 + y^2 = v_{0x}^2 t^2 + v_{0y}^2 t^2 - v_{0y} g t^3 + \frac{1}{4}g^2t^4
\Rightarrow v_0^2 t^2 + \frac{1}{4} g^2 t^4 &gt; v_{0y} g t^3
\Leftrightarrow \frac{v_0^2}{t}+ \frac{1}{4} g^2 t &gt; v_0 cos(\alpha) g
\Leftrightarrow \frac{v_0}{g t}+ \frac{g t}{4v_0} &gt; cos(\alpha)
\Leftrightarrow \alpha &lt; arcos \left(\frac{v_0}{g t}+ \frac{g t}{4v_0}\right)

and that has to hold for every v0, alpha and t
not sure about this tho :P

 

[blindside]

Its amazing how many mistakes you can make typing something like this out in TeX. This question is from a text I use all the time, and it always explicitly states 'in terms of var1, var2...' if the solution is general. It's looking for a numerical angle :/ This leads me to believe I'm looking at this the wrong way.

 

[psi^]

yeah :)
i am new to TeX too :'( next time i'll do it on paper first! LoL

 

[blindside]

This is starting to get extremely frustrating...

I find the same solution approaching the problem a different way.

\vec{r}(t) = (v_{0}\cos{\alpha_{0}}t)\mathbf{\hat{i}} + (v_{0}\sin{\alpha_{0}}t-\frac{1}{2}gt^2)\mathbf{\hat{j}}
\vec{v}(t) = (v_{0}\cos{\alpha_{0}})\mathbf{\hat{i}} + (v_{0}\sin{\alpha_{0}}-gt)\mathbf{\hat{j}}

Therefore:

\vec{r}(t)\cdot\vec{v}(t) = |\vec{r}||\vec{v}|\cos{\phi} = t(v_{0}^2-\frac{3}{2}v_{0}gt\sin{\alpha_{0} + \frac{1}{2}g^2t^2)

Projectile is moving away from the origin if the angle \vec{v} makes with \vec{r} is \pm90\,^{\circ}, i.e. \cos{\phi} &gt; 0 since |\vec{r}|, |\vec{v}| are positive. Since t, v, g, are all known to be positive...

g^2t^2-3gtv_{0}\sin{\alpha_{0}}+2v_{0}^2 &gt; 0

3gtv_{0}\sin{\alpha_{0}} &lt; 2v_{0}^2+g^2t^2

\sin{\alpha_{0}} &lt; \frac{2v_{0}}{3gt}+\frac{gt}{3v_{0}}

 

[physixguru]

https://www.physicsforums.com/showthread.php?t=189307

 

[blindside]

Cheers, got it. Shame that thread didn't come up when I searched yesterday :(