Why Is the Trajectory of a Projectile a Parabola?

The trajectory is a parabola because, without fluid resistance, a projectile has constant horizontal velocity but constant vertical acceleration. Eliminating time from these two motions produces a quadratic equation relating vertical displacement to horizontal displacement.

The Reasoning

This is part of A.1 Kinematics for both SL and HL. Consider a projectile launched from the origin with initial speed $u$ at angle $\theta$ above the horizontal.

These equations follow from resolving the initial velocity into perpendicular components. Gravity acts only vertically, so horizontal displacement is proportional to time, whereas vertical displacement contains a term proportional to the square of time. The same time variable links the two components.

Rearrange the horizontal equation to make time the subject:

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

Substitute this expression into the vertical equation:

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

Simplifying gives:

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

This has the form $y=ax+bx^2$, which is a quadratic relationship. Its graph is therefore a downward-opening parabola because the coefficient of $x^2$ is negative.

At equal horizontal intervals, time increases by equal amounts, but downward displacement due to gravity grows as $t^2$. This changing vertical displacement produces the curvature; constant horizontal velocity alone would produce a straight line. The path is symmetric about its maximum height.

This result assumes a uniform gravitational field, negligible fluid resistance, and a non-vertical launch. With air resistance, horizontal velocity decreases and the trajectory is not exactly parabolic.

A common misconception is that acceleration becomes zero at the highest point. Only vertical velocity is zero there; acceleration remains $g$ downward throughout the flight.

Exam Technique

For an IB Show question, write both component equations, eliminate $t$, and simplify until the quadratic form appears. State the assumptions explicitly and remember that horizontal and vertical motions share the same time but are analysed independently.