Year 12 Maths Extension 1 (2027) Vectors

Projectile motion – equation of path

Q: What is the path of a projectile?

A downward parabola, y equals x tan theta minus g x squared over two V squared cos squared theta.

Q: How do you get the Cartesian equation?

Eliminate t using t equals x over V cos theta and substitute into y.

Q: How do you find the range?

Set y equals zero and solve for the non-zero x.

Q: How do you find the greatest height?

Find the vertex of the parabola, where dy by dx equals zero.

20 practice questions 2 video lessons Theory + worked examples

Understand the equation of path for projectile motion in NSW Year 12 Mathematics Extension 1. A projectile moves under gravity alone, so its horizontal and vertical motion can be modelled separately and then combined.

You will learn to resolve the initial velocity into components, write the parametric equations of motion, and eliminate time to find the Cartesian equation of the path — the parabola that describes a real trajectory in the Extension 1 course.

Practice 20 questions

Practice questions

Every question with a fully worked solution.

Start practising

Watch 2 video(s)

Equation of Trajectory of Projectiles - Mechanics Year 13 A-Level Watch
The path of a projectile. Parametric equations Watch

Theory

A projectile splits into horizontal (constant velocity) and vertical (acceleration $- g$ ) motion, giving $x = V \cos θ t$ , $y = V \sin θ t - \frac{1}{2} g t^{2}$ and the parabolic path $y = x \tan θ - \frac{g x^{2}}{2 V^{2} \cos^{2} θ}$ . This NSW Year 12 Mathematics Extension 1 topic is NESA outcome ME1-12-02.

A projectile moves under gravity alone. Splitting the motion into a horizontal part (constant velocity) and a vertical part (constant acceleration $- g$ ) gives parametric equations; eliminating $t$ gives the parabolic path.

Projected from $O$ with speed $V$ at angle $θ$ , with acceleration $- g j$ :

x = V \cos θ t, y = V \sin θ t - \frac{1}{2} g t^{2} .

Eliminating $t = \frac{x}{V \cos θ}$ gives the Cartesian path $y = x \tan θ - \frac{g x^{2}}{2 V^{2} \cos^{2} θ}$ , a downward parabola.

NESA link. Part of the Year 12 Introduction to vectors focus area, outcome ME1-12-02 ("operates with 2D and 3D vectors and uses 2D vectors to solve problems involving motion in two dimensions") with MAO-WM-01. The full projectile treatment appears on the next page.

Launched at speed

V

, angle

θ

; the path is a downward parabola.

V

splits into

V \cos θ

(horizontal) and

V \sin θ

(vertical).

x = V \cos θ t, y = V \sin θ t - \frac{1}{2} g t^{2} .

x = V cos(theta) t; y = V sin(theta) t - (1/2) g t^2

y = x \tan θ - \frac{g x^{2}}{2 V^{2} \cos^{2} θ} (eliminate t = \frac{x}{V \cos θ}) .

path: y = x tan(theta) - g x^2 / (2 V^2 cos^2 theta)

Two separate motions. Horizontal velocity is constant ( $\ddot{x} = 0$ ); vertical acceleration is $- g$ . The path is always a downward parabola.

How to find the path

Write the components: $x = V \cos θ t$ , $y = V \sin θ t - \frac{1}{2} g t^{2}$ .
Make $t$ the subject of the $x$ -equation: $t = \frac{x}{V \cos θ}$ .
Substitute into $y$ to get the Cartesian path.
Use the path to find range ( $y = 0$ ) and greatest height (vertex).

Example 1 — From velocity

A particle is projected from

O

with velocity

8 i + 6 j

(take

g = 10

). Find the Cartesian path.

Solution

$x = 8 t$ , $y = 6 t - 5 t^{2}$ , so $t = \frac{x}{8}$ .

y

=

6 \cdot \frac{x}{8} - 5 {(\frac{x}{8})}^{2} = \frac{3 x}{4} - \frac{5 x^{2}}{64}

y = 3x/4 - 5x^2/64

$y = \frac{3 x}{4} - \frac{5 x^{2}}{64}$ .

Example 2 — Range

A projectile follows

y = x - \frac{x^{2}}{50}

. Find its horizontal range.

Solution

Set $y = 0$ .

x (1 - \frac{x}{50})

=

0 \Rightarrow x = 0 or 50

range = 50

The range is $50$ .

Example 3 — Greatest height

A projectile follows

y = 3 x - \frac{x^{2}}{12}

. Find the greatest height.

Solution

The vertex is where $\frac{d y}{d x} = 0$ .

$3 - \frac{x}{6}$	$=$	$0 \Rightarrow x = 18$
$y$	$=$	$54 - 27 = 27$

greatest height = 27

Greatest height $= 27$ .

Example 4 — Derive the path

With

x = V \cos θ t

and

y = V \sin θ t - \frac{1}{2} g t^{2}

, show

y = x \tan θ - \frac{g x^{2}}{2 V^{2} \cos^{2} θ}

Solution

$t = \frac{x}{V \cos θ}$ ; substitute.

$y$	$=$	$V \sin θ \cdot \frac{x}{V \cos θ} - \frac{1}{2} g {(\frac{x}{V \cos θ})}^{2}$
	$=$	$x \tan θ - \frac{g x^{2}}{2 V^{2} \cos^{2} θ}$

substituting gives y = x tan(theta) - g x^2/(2 V^2 cos^2 theta)

As required.

Common pitfalls

Two separate motions. Horizontal has constant velocity; vertical has acceleration

- g

— don't mix them.

Eliminating $t$ . Use

t = \frac{x}{V \cos θ}

from the horizontal equation.

Sign of gravity. The vertical acceleration is

- g

(downward).

The path is a parabola. It always opens downward.

Frequently asked questions

What is the path of a projectile?

A downward parabola $y = x \tan θ - \frac{g x^{2}}{2 V^{2} \cos^{2} θ}$ .

How do you get the Cartesian equation?

Eliminate $t$ using $t = \frac{x}{V \cos θ}$ and substitute into $y$ .

Why split the motion into components?

Horizontal velocity is constant and vertical acceleration is $- g$ , so each is simple.

How do you find the range?

Set $y = 0$ and solve for the non-zero $x$ .

How do you find the greatest height?

Find the vertex of the parabola, where $\frac{d y}{d x} = 0$ .

← Previous subtopic

Motion in vector form 2D (velocity/accel, relative velocity, crosswind)

Next subtopic →

Projectile motion – flight / time equations

Projectile motion – equation of path

📖 Theory

How to find the path

Common pitfalls

Frequently asked questions

Theory