Equation of Trajectory

What this equation describes

The position equations $x = u\cos\theta \cdot t$ and $y = u\sin\theta \cdot t - \frac{1}{2}gt^2$ describe where the projectile is at each instant in time. But what is the shape of the path itself — the curve traced in space, independent of time?

The equation of trajectory answers this. It gives $y$ directly as a function of $x$, with time eliminated:

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

This is the equation of a downward-opening parabola.

Derivation

Start with the parametric position equations:

$x = u\cos\theta \cdot t \quad \text{...(1)}$

$y = u\sin\theta \cdot t - \frac{1}{2}gt^2 \quad \text{...(2)}$

From equation (1), express $t$ in terms of $x$:

$t = \frac{x}{u\cos\theta}$

Substitute into equation (2):

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

$y = \frac{\sin\theta}{\cos\theta} \cdot x - \frac{g}{2u^2\cos^2\theta} \cdot x^2$

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

Why this is a parabola

The equation has the form $y = Ax - Bx^2$ where:

$A = \tan\theta, \qquad B = \frac{g}{2u^2\cos^2\theta}$

This is a quadratic in $x$ with a negative coefficient on $x^2$ (since $B > 0$). Any equation of the form $y = Ax - Bx^2$ is a downward-opening parabola. The path of every projectile (under constant gravity, no air resistance) is a parabola — this is a fundamental result.

Key features of the trajectory

$y$-intercept: At $x = 0$: $y = 0$ — the parabola passes through the origin (launch point).

Where it crosses the ground again: Set $y = 0$:

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

$0 = x\left(\tan\theta - \frac{gx}{2u^2\cos^2\theta}\right)$

Solutions: $x = 0$ (launch) and $x = \frac{2u^2\sin\theta\cos\theta}{g} = \frac{u^2\sin2\theta}{g} = R$ (range). Consistent with the range formula.

Maximum height: At the vertex of the parabola, $x = \frac{R}{2}$:

$y_{\max} = \frac{R}{2}\tan\theta - \frac{g(R/2)^2}{2u^2\cos^2\theta} = \frac{u^2\sin^2\theta}{2g} = H$

Consistent with the maximum height formula.

Alternate form using $\sec^2\theta$

Since $\frac{1}{\cos^2\theta} = \sec^2\theta = 1 + \tan^2\theta$:

$y = x\tan\theta - \frac{gx^2}{2u^2}(1 + \tan^2\theta)$

This form is useful when $\tan\theta$ is given directly in a problem rather than $\theta$.

What changes the shape

• Larger $u$ → wider, flatter parabola (body travels further before gravity pulls it down)
• Larger $g$ → narrower, steeper parabola (gravity pulls it down faster)
• Larger $\theta$ → steeper initial slope (but range may decrease if $\theta > 45°$)