Trying to find angle at which projectile was fired.

So let's say I have a parabolic function that describes the displacement of some projectile without air resistance. Let's say it's

$$y=-4.9x^2+V_0x.$$

I want to know at what angle the projectile was fired. I notice that

$$\tan \theta_0=f'(x_0)$$ so the angle should be

$$\theta_0 = \arctan(f'(0)).$$ or

$$\theta_0 = \arctan(V_0).$$

Is this correct? I can't work out why it wouldn't be, but it doesn't look right when I plot curves.

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If you intend -9.8 to be the acceleration of gravity, the altitude it contributes should be $\frac {-9.8}2 x^2$ (where you are using $x$ for time, not horizontal distance) –  Ross Millikan Aug 28 '12 at 19:43
oops! Yep. I forgot. :) –  Korgan Rivera Aug 28 '12 at 20:03
As the others say, if $x$ is time you are working in one dimension and there is no $\theta_0$ If you are in two dimensions, you need $x$ and $t$. –  Ross Millikan Aug 28 '12 at 20:10
@Ross: If the OP meant $x$ to be time, you're correct. However, it seems more likely to me that they really did mean $x$ to be horizontal position, in which case $\theta=\arctan f'(x_0)$ does, in fact, give the angle of the trajectory (up to sign, anyway) at position $(x_0,f(x_0))$. –  Ilmari Karonen Aug 29 '12 at 1:11
@IlmariKaronen: but then the first equation doesn't work in units at all. The acceleration of gravity shouldn't multiply distance^2 to get distance. OP edited in response to my first comment, which seems to confirm that $x$ is time. –  Ross Millikan Aug 29 '12 at 1:52

It helps to define your terms. $x$ looks like a distance, but you are using it as time in your first equation. I would start with let $t$ be the time after the projectile is launched. Let $\vec {V_0}=V_{0x} \hat i+V_{0y} \hat j$ be the initial velocity. We have $V_{0x}=|V_0| \cos \theta_0, V_{0y}=|V_0| \sin \theta_0$. Then the equation of motion is $y=-\frac {9.8}2 t^2+V_{0x}t$. It is true that $\theta_0=\arctan \frac {V_{0y}}{V_{0x}}$

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From what you are saying, it looks like you are thinking of $f(x)$ as a real valued function, and in effect you are only considering linear motion. (This would be fine, if the cannon were firing straight up or straight down or straight up.)

However, you are interested in knowing the interesting two-dimensional trajectories, and this means that $V_0$ is a 2-D vector pointing in the direction of the initial trajectory, and that the coefficient of $x^2$ actually accompanies a vector which is pointing straight down (to show the contribution of gravity.)

Once you find what $V_0$ is, you can then compute its angle with respect to horizontal, and have your answer.

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As you only have one equation to denote the position I assume you are working in one dimension. If you are working in one dimension, the initial angle is $0$ or $\frac{\pi}{2}$, if your $y$ denotes the height of the position. For a 2 dimensional trajectory you need two equations to denote the position of a projectile. If you shoot a projectile at speed $v_0$ with an angle $\alpha$, the position $(x,y)$ is $$x=x_0+v_0t\cos\alpha$$ $$y=y_0+v_0t\sin\alpha+a{t}^{2}$$ where $(x_0,y_0)$ is the initial position, $a$ is the gravity and $t$ is the time. If you know $v_x=v_0\cos\alpha$ and $v_y=v_0\sin\alpha$, the initial angle $\alpha$ would be $\arctan\frac{v_y}{v_x}$. You could do a function that denotes the angle in function of time $\alpha(t)=\arctan\frac{v_0\sin\alpha+2at}{v_0\cos\alpha}$. The angle in the vertex of the parabola that describes the projectile should be $0^o$, as $v_y$ should be zero. So $\alpha(\frac{-v_0\sin\alpha}{2a})=0$. If $x_0=0$ and $y_0=0$, then $\alpha(\frac{-v_0\sin\alpha}{a})=-\arctan\frac{v_y}{v_x}$. Summarizing, with your equation, the trajectory is vertical so the angle $\alpha$ would be $\lim_{x\to 0}\arctan\frac{V_0}{x.V_0} = \frac{\pi}{2}$ as $x$ does not change.

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Yes, it's correct.

In fact, it would be correct, by definition, for any trajectory: if the path of an object in the $(x,y)$ plane can be written as $y = f(x)$ for some differentiable function $f$, and the object is moving along the path in the direction of increasing $x$, then the direction the object moves in (expressed as an angle from the horizontal) when it crosses the line $x = x_0$ is given by $\arctan f'(x_0)$.

As for why your results might not look like what you expect, well, the first question that comes to mind is whether your $\arctan$ function returns degrees or radians. Being off by a factor of $180/\pi$ would certainly make the results look odd.

If that's not the issue, then the next thing I'd check is whether you actually calculated $f'$ correctly.

Ps. I'm assuming that you did, in fact, mean $x$ to be horizontal position rather than time. However, if so, your definition of $f$ is somewhat odd. Generally, if a projectile is fired at time $t_0$ from position $(x_0,y_0)$ with speed $v_0$ and angle $\theta$, and is pulled down by gravity with acceleration $g$, the equations of motions are

\begin{aligned} x &= x_0 + v_0 \cos(\theta)\, t \\ y &= y_0 + v_0 \sin(\theta)\, t - \frac12 g t^2, \end{aligned}

the first of which we may solve for $t$ and substitute into the second to get the trajectory

\begin{aligned} y &= y_0 + v_0 \sin(\theta) \frac{x-x_0}{v_0 \cos(\theta)} - \frac12 g \left(\frac{x-x_0}{v_0 \cos(\theta)}\right)^2 \\ &= y_0 + \tan(\theta) (x-x_0) - \frac{g}{2v_0^2\cos^2(\theta)}(x-x_0)^2. \end{aligned}

Assuming that $x_0 = y_0 = 0$, and comparing the result with your original equation, we see that the coefficient $\tan(\theta)$ corresponds to your $V_0$, which is thus not a velocity but a slope, while the coefficient $-9.8$ (or $-4.9$ as you edited it) corresponds not to $-g$ or $-g/2$, but to $-g/(2v_0^2\cos^2(\theta))$. Only in the special case where the horizontal velocity component $v_0\cos(\theta) = 1$ are the latter two expressions equal (as they should be, given that in that case $t = x$).

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