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.

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.
