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August 9th, 2014, 07:20 PM   #1
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Math Focus: Calculus, Differential Geometry, Physics, Topology
Projectile Motion (only diff calc required!)

Hey guys,

Here's a really neat projectile motion problem that I grabbed from my University Physics textbook.

What is the greatest angle, with respect to the horizontal, at which a projectile can be thrown such that it is always getting further away from you?

That is to say, imagine a vector, \(\vec{r}\), the position vector of the projectile, with respect to you. Find the maximum angle above the horizontal that the projectile could've been thrown if the magnitude of \(\vec{r}\) never decreases.


SPOILERS:
Here's my work so far, during which I've found the answer in the back of the book. I am confused, because my function for the derivative does have a root, and it's positive, and depends on the initial velocity of the projectile. What is the meaning of this root?

If you haven't tried the problem, I encourage you not to peak, but here's my work: https://dl.dropboxusercontent.com/u/...projectile.pdf
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August 9th, 2014, 08:30 PM   #2
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The question doesn't stipulate that the angle must be above the horizontal (although I'd be amazed if it weren't).
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August 10th, 2014, 07:04 AM   #3
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Quote:
Originally Posted by v8archie View Post
The question doesn't stipulate that the angle must be above the horizontal (although I'd be amazed if it weren't).
Any angle below the horizontal will work, if you imagine the projectile can keep falling below the horizontal. (mathematically, it can) The question asks for the largest angle, so those negative angles end up included!
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August 10th, 2014, 06:16 PM   #4
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I have the following:
  • Initial speed: $v_0$
  • Angle of throw: $\theta$
  • Position at time $t$: $P(x(t),y(t))$
  • Distance from start: $d(t) = \sqrt{x^2(t) + y^2(t)}$

Since the distance is always positive we will, for convenience, use \begin{align*}
s(t) &= d^2(t) = x^2(t) + y^2(t) \\
\dot{s}(t) &= 2x\dot{x} + 2y\dot{y} \\[12pt]
\end{align*}
For this calculation we will then need
\begin{align*}
\ddot{x}(t) &= 0 \\
\dot{x}(t) &= v_0\cos\theta \\
x(t) &= v_0t\cos\theta\\[12pt]
\ddot{y}(t) &= -g \\
\dot{y}(t) &= v_0\sin\theta-gt \\
y(t) &= v_0t\sin\theta - \tfrac12 gt^2 \\
\end{align*}
We are interested in \begin{align*}\dot{s}(t) &\ge 0 \\
x\dot{x} + y\dot{y} &\ge 0 \\
\left(v_0t\sin\theta - \tfrac12 gt^2\right)\left(v_0\sin\theta-gt\right)+v_0t\cos\theta \cdot v_0\cos\theta &\ge 0 \\
v_0^2t\sin^2\theta - \tfrac32gt^2v_0\sin\theta + \tfrac12g^2t^3 + v_0^2t\cos^2\theta &\ge 0 \\
v_0^2t - \tfrac32gt^2v_0\sin\theta + \tfrac12g^2t^3 &\ge 0 \\
\tfrac12t\left(g^2t^2 - 3gv_0t\sin\theta +2v_0^2\right) &\ge 0 \\
\end{align*}

For this to be true for all $t \ge 0$ we need $(g^2t^2 - 3gv_0t\sin\theta +2v_0^2)$ to have no positive real roots. Since $\tfrac{3}{2g}v_0 \gt 0$, we either require $\sin\theta \le 0 \implies 0 \le \theta \le -\frac{\pi}{2}$ (a downward throw) or for the quadratic to have no real roots:
\begin{align*} (-3gv_0\sin\theta)^2 - 8g^2v_0^2 &\lt 0 \\
9g^2v_0^2\sin^2\theta &\lt 8g^2v_0^2 \\
\sin^2\theta &\lt \tfrac{8}{9} \\
\sin\theta &\lt \tfrac{2}{3}\sqrt2 \\
\end{align*}
Where the negative solution is ignored because we have already covered the case with negative $\theta$.
Thanks from topsquark and russphelan

Last edited by v8archie; August 10th, 2014 at 06:19 PM.
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August 11th, 2014, 06:21 PM   #5
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I think that the fact that $\theta$ is independent of $g$ is rather counter-intuitive. Independence with the initial velocity seems straightforward, as velocity simply scales the resultant graph equally in the $x$ and $y$ directions. But that can't be the case for $g$.

The best reason that I can come up with is that, once launched, the angle made by the velocity with the horizontal is always decreasing. Moreover, the angle at $y=0$ on the way down is the same (up to a change of sign) as on the way up. So while other trajectories will reach the same angle with the horizontal, those starting with a smaller angle only attain it when the ball has already passed the starting point on the way down.
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