# Why is the angle of impact complementary to the angle of launch in the simple equations for the range of a projectile?

I'm using the standard equation for the range of a projectile:

\begin{align} d &= \frac{v\ \text{cos}\theta}{g} \left( v\ \text{sin}\theta + \sqrt{v^2\ \text{sin}^2\theta + 2gy_0}\right) \end{align}

where $d$ is the horizontal distance traveled, along with the associated equation for the angle $\psi$ of the projectile's trajectory in the moment before impact:

\begin{align} \text{tan}\psi &= \frac{\sqrt{v^2\ \text{sin}^2\theta + 2gy_0}} {v\ \text{cos}\theta} \end{align}

In the section on angle of impact, wiki provides a derivation for the fact that $\theta + \psi = 90^\circ$, and I failed when attempting to derive this equation symbolically, possibly because I don't remember a useful trig identity. I wasn't entirely clear on some of the substitutions the wiki article made, so I made my own attempt:

\begin{align} \text{tan}\psi &= \frac{\sqrt{v^2\ \text{sin}^2\theta + 2gy_0}} {v\ \text{cos}\theta} \\ \text{tan}^2\psi &= \frac{v^2\ \text{sin}^2\theta + 2gy_0} {v^2\ \text{cos}^2\theta} \\ &= \text{tan}^2\theta + \frac{2gy_0}{v^2\ \text{cos}^2\theta} \\ \text{tan}^2\psi - \text{tan}^2\theta &= \frac{2gy_0}{v^2\ \text{cos}^2\theta} \\ (\text{tan}\psi + \text{tan}\theta)(\text{tan}\psi - \text{tan}\theta) &= \frac{2gy_0}{v^2\ \text{cos}^2\theta} \end{align}

At this point, I was tempted to use the trig identities: \begin{align} \text{tan}(\psi + \theta) &= \frac{\text{tan}\psi + \text{tan}\theta}{1-\text{tan}\psi\ \text{tan}\theta} \\ \text{tan}(\psi - \theta) &= \frac{\text{tan}\psi - \text{tan}\theta}{1+\text{tan}\psi\ \text{tan}\theta} \end{align}

but making that substitution didn't seem to help anything, so I checked that I could at least get this result numerically. No such luck, however. As an example, I used $g = 9.81\ \text{m/s}^2, \theta=30^\circ,v=10 \ \text{m/s},$ and $y_0=3 \ \text{m}.$ This gives me:

\begin{align} \text{tan}\psi &= \frac{\sqrt{10^2\ \text{sin}^2 30 + 2(-9.81)(3)}} {10\ \text{cos}30} \\ &= 1.05742 \\ \therefore \ \ \ \psi &= 46.5985^\circ \\ \end{align}

I don't see any errors in the exceedingly simple Matlab code I used to calculate this:

g = 9.81;
theta = 30;
v = 10;
y0 = 3;
psiAng = atand(sqrt((v*sind(theta))^2 + 2*g*y0)/(v*cosd(theta)));


If I assume an initial height $y_0 = 0$ but keep the other values, I get $\psi \approx 30^\circ = \theta$, which doesn't satisfy this property either.

Is $\theta + \psi = 90^\circ$, or have I committed one or more errors in the above calculations? I'm ignoring all other factors not present in these equations, e.g. air resistance.

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Are you taking into account the fact that the article stipulates "for maximum range?" That is, $\theta + \psi = 90^\circ$ only if $\theta$ is chosen so as to maximize $d$. – Chris White May 14 '13 at 2:27
@ChrisWhite I wasn't entirely sure if that statement ("for maximum range") applied to the remainder of the proof the article presented, but looking again, it does read that way. I'm not sure why I had a brain lapse and didn't read it that way at first. I guess that's it, then. – John Bensin May 14 '13 at 2:31
That does seem to be a very silly little section of wikipedia. First, you assume that $\theta = 45^{\circ}$; then you prove that $2\theta = 90^{\circ}$... – Mike May 14 '13 at 2:37
@Mike Could it be that $\theta + \psi = 90^\circ$ whenever maximum horizontal distance is the case, even if the initial and final heights are different? Then the result is a little less trivial. Even so, the whole article is rather cumbersome and obscure. – Chris White May 14 '13 at 3:22
@JohnBensin If that solves the issue, feel free to post it as an answer. – Chris White May 14 '13 at 3:24