NSWMaths Extension 1Syllabus dot point

How do we model projectile motion in two dimensions, and what quantities (range, maximum height, time of flight) can we extract?

Model projectile motion in two dimensions using parametric equations and find range, maximum height, time of flight and trajectory equation

A focused answer to the HSC Maths Extension 1 dot point on projectile motion. The parametric equations for position, velocity and acceleration under gravity, the Cartesian trajectory equation, and standard quantities (range, maximum height, time of flight), with worked examples.

Generated by Claude OpusReviewed by Better Tuition Academy9 min answerUpdated 2026-05-20

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What this dot point is asking

NESA wants you to model a projectile launched from ground level (or some height) with initial speed and angle, write down the parametric position equations, find the trajectory equation, and compute the standard quantities: range, time of flight, maximum height, and velocity at a given time.

The answer

The model

A projectile launched from the origin with initial speed $V$ at angle $\alpha$ above the horizontal, with gravitational acceleration $g$ downward and no air resistance:

Velocity components:

\dot x(t) = V \cos \alpha, \qquad \dot y(t) = V \sin \alpha - g t.

Position equations:

x(t) = V \cos \alpha \cdot t, \qquad y(t) = V \sin \alpha \cdot t - \tfrac{1}{2} g t^2.

Acceleration:

\ddot x = 0, \qquad \ddot y = -g.

Horizontal motion is at constant velocity; vertical motion is under constant downward acceleration $g$ .

Standard derived quantities

Time of flight (on horizontal ground): set $y(t) = 0$ .

T = \frac{2 V \sin \alpha}{g}.

Range on horizontal ground: $R = V \cos \alpha \cdot T$ .

R = \frac{V^2 \sin 2 \alpha}{g}.

(Maximum range occurs at $\alpha = 45^\circ$ .)

Maximum height: set $\dot y(t) = 0$ , $t = \frac{V \sin \alpha}{g}$ , then substitute.

H = \frac{V^2 \sin^2 \alpha}{2 g}.

Trajectory (Cartesian) equation

Eliminate $t$ from the position equations. From $x = V \cos \alpha \cdot t$ , $t = \frac{x}{V \cos \alpha}$ . Substitute into $y$ :

y = x \tan \alpha - \frac{g x^2}{2 V^2 \cos^2 \alpha} = x \tan \alpha - \frac{g x^2 (1 + \tan^2 \alpha)}{2 V^2}.

This is a downward parabola.

Launched from a height

If the projectile starts at height $h$ , then $y(t) = h + V \sin \alpha \cdot t - \tfrac{1}{2} g t^2$ . The time of flight is now the larger root of this quadratic in $t$ ; the range is no longer the symmetric formula.

Velocity speed and direction

Magnitude of velocity at time $t$ :

v(t) = \sqrt{(\dot x)^2 + (\dot y)^2}.

Direction (above horizontal):

\theta(t) = \arctan \frac{\dot y}{\dot x}.

Note that the projectile speed is minimum at the top of the trajectory (where $\dot y = 0$ ) and equals $V \cos \alpha$ there.

Range on inclined ground

For ground inclined at angle $\beta$ below horizontal at launch (downhill) or above (uphill), the time of flight and range require setting $y(t) = -x(t) \tan \beta$ (for downhill) or solving the relevant intersection. This is a more involved application and shows up in harder Section II questions.

Worked example

Basic range

A ball is kicked from ground level at $20$ m/s at $30^\circ$ . Find the range. Take $g = 9.8$ m/s $^2$ .

$R = \frac{V^2 \sin 2 \alpha}{g} = \frac{400 \cdot \sin 60^\circ}{9.8} = \frac{400 \cdot 0.866}{9.8} \approx 35.3$ m.

Maximum height

Same launch. $H = \frac{V^2 \sin^2 \alpha}{2 g} = \frac{400 \cdot 0.25}{19.6} \approx 5.10$ m.

Time at a given height

The ball reaches what height at $t = 1.5$ s?

$y(1.5) = 20 \sin 30^\circ \cdot 1.5 - \tfrac{1}{2} \cdot 9.8 \cdot 1.5^2 = 15 - 11.025 = 3.975$ m. About $4.0$ m.

Trajectory equation

Find the Cartesian trajectory for a ball at $25$ m/s at $40^\circ$ .

$y = x \tan 40^\circ - \frac{9.8 \, x^2}{2 \cdot 625 \cdot \cos^2 40^\circ} \approx 0.839 x - 0.0134 \, x^2$ .

Velocity at landing

A ball is thrown at $V = 15$ m/s at $\alpha = 60^\circ$ . Find the speed at landing.

By symmetry on level ground, the speed at landing equals the speed at launch: $15$ m/s. The angle below horizontal at landing also equals $60^\circ$ .

Maximum-range angle on level ground

A footballer kicks the ball at fixed speed. What launch angle gives maximum range?

$R = \frac{V^2 \sin 2 \alpha}{g}$ is maximised when $\sin 2 \alpha = 1$ , that is $2 \alpha = 90^\circ$ , so $\alpha = 45^\circ$ .

Common traps

Sign of gravity: With upward positive, gravitational acceleration is $-g$ . Mixing the sign in the position equation gives a wrong shape.
Confusing initial speed with component: IMATH_1 is the horizontal component, $V \sin \alpha$ is the vertical. Mixing them swaps the trajectory.
Wrong angle in $\sin 2 \alpha$: The range formula uses $\sin 2 \alpha$ , not $\sin^2 \alpha$ . The double angle is from the product $\sin \alpha \cos \alpha = \frac{1}{2} \sin 2 \alpha$ .
Forgetting starting height: Many HSC questions launch from a cliff or table. The position equation needs the offset $h$ for $y(0)$ .
Mixing units: Use SI ( $g = 9.8$ m/s $^2$ or sometimes $g = 10$ m/s $^2$ if the question allows the approximation). State $g$ explicitly and check the question's value.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2022 HSC Q214 marksA particle is projected from ground level at angle

\alpha

above the horizontal with initial speed

V

. Assuming acceleration due to gravity is

g

, derive expressions for the range

R

on horizontal ground and the maximum height

H

Show worked answer →

Components of initial velocity: $V_x = V \cos \alpha$ , $V_y = V \sin \alpha$ .

Position equations (with the origin at launch and upward positive):

$x(t) = V \cos \alpha \cdot t$

$y(t) = V \sin \alpha \cdot t - \frac{1}{2} g t^2$

Range: set $y(t) = 0$ to find time of flight. $t = 0$ (launch) or $t = \frac{2 V \sin \alpha}{g}$ .

$R = V \cos \alpha \cdot \frac{2 V \sin \alpha}{g} = \frac{2 V^2 \sin \alpha \cos \alpha}{g} = \frac{V^2 \sin 2 \alpha}{g}$ .

Maximum height: set $\dot y(t) = V \sin \alpha - g t = 0$ , so $t = \frac{V \sin \alpha}{g}$ .

$H = V \sin \alpha \cdot \frac{V \sin \alpha}{g} - \frac{1}{2} g \cdot \left( \frac{V \sin \alpha}{g} \right)^2 = \frac{V^2 \sin^2 \alpha}{g} - \frac{V^2 \sin^2 \alpha}{2 g} = \frac{V^2 \sin^2 \alpha}{2 g}$ .

Markers reward stating the equations of motion, finding times by setting position or velocity components to zero, and the standard expressions for $R$ and $H$ .