Cartesian equation of a trajectory

The two equations of motion each contain the time $t$. The trajectory equation is what you get when you eliminate $t$ to leave a direct relationship between $y$ and $x$ — the shape of the path, independent of how fast the clock is running.

Step 1 — make $t$ the subject of the horizontal equation: $x = V\cos\theta\,t \;\Rightarrow\; t = \frac{x}{V\cos\theta}$

Step 2 — substitute into the vertical equation $y = V\sin\theta\,t - \tfrac12 g t^2$: $y = V\sin\theta\left(\frac{x}{V\cos\theta}\right) - \tfrac12 g\left(\frac{x}{V\cos\theta}\right)^2$

Step 3 — simplify each term:

• First term: $\dfrac{V\sin\theta}{V\cos\theta}\,x = x\tan\theta$.
• Second term: $\dfrac{g x^2}{2V^2\cos^2\theta}$.

$\boxed{\,y = x\tan\theta - \frac{g x^2}{2V^2\cos^2\theta}\,}$

This is the Cartesian equation of the trajectory, and it is given in MF19. But a "show that" question wants the full derivation above — every step earns a mark.

Cambridge tip. The first term simplifying to $x\tan\theta$ is the most reassuring checkpoint; if you don't see $\tan\theta$ appear, recheck your $\dfrac{\sin}{\cos}$.

Look at the trajectory equation with $V$ and $\theta$ fixed: $y = (\tan\theta)\,x - \left(\frac{g}{2V^2\cos^2\theta}\right)x^2$

This has the form $y = bx - ax^2$ with $a > 0$ — a parabola opening downward. So the projectile's path is always a parabola (within the no-air-resistance model). The two zeros of $y$ (where $y = 0$) are at $x = 0$ (launch) and at $x = R$ (the range on level ground).

Reading features off the parabola:

• The launch direction is the gradient at $x = 0$: differentiating, $\dfrac{dy}{dx}\big|_{x=0} = \tan\theta$, confirming the launch angle.
• The greatest height is the vertex of the parabola.
• The range on level ground is the non-zero root, $x = R$.

Cambridge tip. Recognising "$y$ quadratic in $x$ $\Rightarrow$ parabola" is sometimes worth an explicit mark; if asked to "describe the path", the word parabola is the key term.

A very common Paper 3 task gives you a point the projectile passes through (and the speed, or the angle) and asks for the missing quantity. The trajectory equation is the tool.

Rewrite using $\sec^2\theta = 1 + \tan^2\theta$. Since $\dfrac{1}{\cos^2\theta} = \sec^2\theta = 1 + \tan^2\theta$, the trajectory equation becomes: $y = x\tan\theta - \frac{g x^2}{2V^2}\left(1 + \tan^2\theta\right)$

Finding the angle $\theta$ (speed $V$ known). Substitute the known $x$, $y$, $V$, $g$. The result is a quadratic in $\tan\theta$: $\frac{g x^2}{2V^2}\tan^2\theta - x\tan\theta + \left(y + \frac{g x^2}{2V^2}\right) = 0$ Solve for $\tan\theta$ — there are usually two solutions (a low trajectory and a high one), both valid unless the question restricts you.

Finding the speed $V$ (angle $\theta$ known). Substitute the known $x$, $y$, $\theta$ into $y = x\tan\theta - \dfrac{g x^2}{2V^2\cos^2\theta}$ and rearrange for $V^2$: $V^2 = \frac{g x^2}{2\cos^2\theta\,(x\tan\theta - y)}$ then take the positive square root.

Cambridge tip. When the quadratic in $\tan\theta$ gives two angles, both are usually correct (the same point can be reached by a flat fast-rising or a steep lobbed shot). State both unless told otherwise — dropping one loses a mark.

Because the trajectory equation is in MF19, examiners often test the derivation instead — a "show that" question. Here the result is given, so quoting it earns nothing; the marks are entirely in the working.

The reliable structure:

1. Write the two equations of motion: $x = V\cos\theta\,t$ and $y = V\sin\theta\,t - \tfrac12 g t^2$.
2. Make $t$ the subject of the first: $t = \dfrac{x}{V\cos\theta}$.
3. Substitute and simplify, showing the $x\tan\theta$ term explicitly.
4. State the final equation and (if asked) convert to the $\tan^2\theta$ form using $\sec^2\theta = 1 + \tan^2\theta$.

Worked-into-a-quadratic 'show that'. Some questions ask you to "show that the angle satisfies $a\tan^2\theta + b\tan\theta + c = 0$" for specific numbers. Substitute the given $x, y, V$ first, then group into the quadratic — clear each fraction so the coefficients are tidy.

Cambridge tip. In a "show that", never skip the substitution of $t$. The most common lost-mark is jumping straight from the two equations to the final answer, which earns the method marks for neither step.