# Projectile Motion and maximum angle

1. Jan 27, 2008

### blindside

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. 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:

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

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:

$$2v_{0}^2-3gtv_{0}\sin{(\alpha_{0})}+g^2t^2 > 0$$
$$\alpha_{0} < \arcsin{\left(\frac{2v_{0}}{3gt}+\frac{gt}{3v_{0}}\right)}$$

Where it's known that:

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

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

Last edited: Jan 27, 2008
2. Jan 27, 2008

### psi^

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

and that has to hold for every v0, alpha and t

Last edited: Jan 27, 2008
3. Jan 27, 2008

### 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.

4. Jan 27, 2008

### psi^

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

5. Jan 28, 2008

### 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. $$\cos{\phi} > 0$$ since $$|\vec{r}|, |\vec{v}|$$ are positive. Since t, v, g, are all known to be positive...

$$g^2t^2-3gtv_{0}\sin{\alpha_{0}}+2v_{0}^2 > 0$$

$$3gtv_{0}\sin{\alpha_{0}} < 2v_{0}^2+g^2t^2$$

$$\sin{\alpha_{0}} < \frac{2v_{0}}{3gt}+\frac{gt}{3v_{0}}$$

Last edited: Jan 28, 2008
6. Jan 28, 2008

### physixguru

7. Jan 28, 2008

### blindside

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

Last edited: Jan 28, 2008