Projectile range symmetry: why 30° and 60° land in the same place

The problem

Launch a projectile at $30°$ and it lands some distance away. Launch it again at the same speed but at $60°$, and it lands in exactly the same spot. This is not a coincidence specific to $30$ and $60$: for any angle $\theta$, launching at $\theta$ and at $90° - \theta$ gives the same range. Why?

The idea

Range depends on launch angle only through $\sin(2\theta)$, and $\sin(2\theta) = \sin(180° - 2\theta) = \sin\big(2(90° - \theta)\big)$, so the sine function's own symmetry does the work. Two complementary angles ($\theta$ and $90° - \theta$) put the same $2\theta$ value, reflected across $90°$, inside the sine, and $\sin$ treats those two inputs identically.

Drag the angle slider below $45°$ and above it. The range markers line up for complementary pairs, but the shape of the flight differs: a low, fast, flat arc and a high, slow, steep one reach the same point by very different paths.

Full proof

Setup. Launch speed $v_0$, launch angle $\theta$, no air resistance, flat ground. Position at time $t$:

$x(t) = v_0 \cos\theta \cdot t, \qquad y(t) = v_0 \sin\theta \cdot t - \tfrac{1}{2} g t^2$

Time of flight. Landing is $y(t) = 0$ for $t > 0$:

$t = 0 \quad\text{or}\quad t = \frac{2 v_0 \sin\theta}{g}$

so the nonzero flight time is $T = \dfrac{2 v_0 \sin\theta}{g}$.

Range. Substituting $T$ into $x(t)$:

$R(\theta) = v_0 \cos\theta \cdot T = \frac{2 v_0^2 \sin\theta\cos\theta}{g} = \frac{v_0^2 \sin(2\theta)}{g}$

using the double-angle identity $2\sin\theta\cos\theta = \sin(2\theta)$.

Symmetry. Compare $R(\theta)$ and $R(90° - \theta)$:

$R(90° - \theta) = \frac{v_0^2 \sin\big(2(90° - \theta)\big)}{g} = \frac{v_0^2 \sin(180° - 2\theta)}{g}$

and since $\sin(180° - x) = \sin(x)$ for any $x$:

$\sin(180° - 2\theta) = \sin(2\theta) \implies R(90° - \theta) = R(\theta) \qquad \blacksquare$

Where it breaks in practice. The error this editorial is filed under, treating equal range as equal trajectory, shows up whenever a problem asks about the flight itself (maximum height, flight time, vertical impact speed) and not just where it lands. Those quantities are not symmetric under $\theta \leftrightarrow 90° - \theta$: maximum height scales with $\sin^2\theta$, which differs for complementary angles unless $\theta = 45°$. Same landing spot, different everything else along the way. The sandbox above shows the whole arc, not just the endpoint, so you can see it.