Projectile Motion and maximum angle

  • Thread starter blindside
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  • #1
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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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Answers and Replies

  • #2
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[tex]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[/tex]
[tex]\Rightarrow v_0^2 t^2 + \frac{1}{4} g^2 t^4 > v_{0y} g t^3[/tex]
[tex]\Leftrightarrow \frac{v_0^2}{t}+ \frac{1}{4} g^2 t > v_0 cos(\alpha) g[/tex]
[tex]\Leftrightarrow \frac{v_0}{g t}+ \frac{g t}{4v_0} > cos(\alpha)[/tex]
[tex]\Leftrightarrow \alpha < arcos \left(\frac{v_0}{g t}+ \frac{g t}{4v_0}\right)[/tex]

and that has to hold for every v0, alpha and t
not sure about this tho :P
 
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  • #3
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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
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yeah :)
i am new to TeX too :'( next time i'll do it on paper first! LoL
 
  • #5
21
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This is starting to get extremely frustrating...

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

[tex]\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}}[/tex]
[tex]\vec{v}(t) = (v_{0}\cos{\alpha_{0}})\mathbf{\hat{i}} + (v_{0}\sin{\alpha_{0}}-gt)\mathbf{\hat{j}}[/tex]

Therefore:

[tex]\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)[/tex]

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

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

[tex]3gtv_{0}\sin{\alpha_{0}} < 2v_{0}^2+g^2t^2[/tex]

[tex]\sin{\alpha_{0}} < \frac{2v_{0}}{3gt}+\frac{gt}{3v_{0}}[/tex]
 
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  • #7
21
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Cheers, got it. Shame that thread didn't come up when I searched yesterday :(
 
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