Projection at Angle from Height — Trajectory

The situation

A body is launched from height $h$ above the ground with speed $u$ at angle $\theta$. Taking the ground as $y = 0$ and the point directly below the launch as $x = 0$, what is the equation of the trajectory?

Derivation

Taking the launch point as the origin temporarily, with upward positive:

$x = u\cos\theta \cdot t$ $y' = u\sin\theta \cdot t - \frac{1}{2}gt^2$

where $y'$ is measured from the launch point.

From the $x$ equation: $t = \frac{x}{u\cos\theta}$

Substituting:

$y' = u\sin\theta \cdot \frac{x}{u\cos\theta} - \frac{1}{2}g\left(\frac{x}{u\cos\theta}\right)^2$

$y' = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}$

But $y'$ is measured from the launch point, and the launch point is at height $h$ above the ground. So the actual height above ground is:

$y = h + y' = h + x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}$

$\boxed{y = h + x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}}$

Relation to the standard trajectory

The standard trajectory (from ground level) is:

$y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}$

The only difference is the addition of $h$ — the launch height shifts the entire parabola upward by $h$.

Where it lands

Set $y = 0$ to find the landing point:

$0 = h + x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}$

$\frac{gx^2}{2u^2\cos^2\theta} - x\tan\theta - h = 0$

This is a quadratic in $x$. The positive root gives the range $R$.

Special case: horizontal projection ($\theta = 0$)

$y = h + 0 - \frac{gx^2}{2u^2} = h - \frac{gx^2}{2u^2}$

The trajectory starts at height $h$ and curves downward. Setting $y = 0$:

$h = \frac{gx^2}{2u^2} \implies x = u\sqrt{\frac{2h}{g}} = R$

Consistent with the horizontal projection range formula.