Projectile motion (QCE Physics Unit 3)

Syllabus dot point

“Solve problems involving projectile motion by resolving the motion into independent horizontal and vertical components, assuming constant downward acceleration due to gravity and negligible air resistance”

10 min readUpdated 22 May 2026

Quick answer

Projectile motion is solved by resolving the launch velocity into independent horizontal and vertical components, applying constant-velocity equations horizontally and constant-acceleration equations vertically (with $g$ down), and linking the two axes through a shared time of flight.

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What this dot point is asking
The answer
How this appears in IA1 and IA2
Examples in context
Try this
Exam-style questions

What this dot point is asking

QCAA wants you to model the motion of a projectile (an object moving only under gravity) by resolving its velocity into independent horizontal and vertical components, then applying the constant-acceleration equations to each axis. The two stated assumptions are constant downward acceleration $g = 9.8 \text{ m/s}^2$ and negligible air resistance. The dot point underpins every projectile calculation in IA1 and EA Paper 2, and is the classic IA2 design (range vs angle).

The answer

A projectile is any object in free flight subject only to gravity. The horizontal and vertical motions are independent, linked only by the shared time of flight.

Resolving the initial velocity

If a projectile is launched with speed $v_0$ at angle $\theta$ above the horizontal:

v_{0x} = v_0 \cos\theta

v_{0y} = v_0 \sin\theta

Horizontal motion

No horizontal force acts (air resistance is ignored), so the horizontal velocity is constant.

x = v_{0x} t

Vertical motion

The only acceleration is gravity, $a_y = -g$ (taking up as positive). Use the constant-acceleration equations:

v_y = v_{0y} - g t

\Delta y = v_{0y} t - \tfrac{1}{2} g t^2

v_y^2 = v_{0y}^2 - 2 g \Delta y

Key features of the trajectory

The path is a parabola. At maximum height $v_y = 0$ , so the rise above the launch point is:

\Delta y_{\max} = \frac{v_{0y}^2}{2g}

For a projectile launched from and landing at the same height, the time of flight is $t = 2 v_{0y} / g$ and the range is:

R = \frac{v_0^2 \sin(2\theta)}{g}

On level ground the range is maximised at $\theta = 45°$ , and complementary launch angles (for example, $30°$ and $60°$ ) give the same range.

If the launch and landing heights differ (release height above the ground, or landing on a raised platform), do not use the level-ground range formula. Set the displacement $\Delta y$ at landing explicitly and solve the vertical equation for $t$ , then use $x = v_{0x} t$ .

Worked example

A ball is kicked from ground level at $v_0 = 20$ m/s at $\theta = 35°$ above horizontal. Find the maximum height, time of flight, and range.

Resolve: $v_{0x} = 20 \cos 35° = 16.38$ m/s, $v_{0y} = 20 \sin 35° = 11.47$ m/s.

Maximum height: $h_{\max} = v_{0y}^2 / (2g) = 11.47^2 / 19.6 = 6.71$ m.

Time of flight: $t = 2 v_{0y} / g = 22.94 / 9.8 = 2.34$ s.

Range: $R = v_{0x} t = 16.38 \times 2.34 = 38.3$ m.

Try it: Projectile motion calculator. Enter launch speed, angle and release height to get the range, max height and trajectory.

How this appears in IA1 and IA2

IA1 data test. Expect either a launch-angle table to interpret (range vs angle data, often missing one value), or a single trajectory with annotated heights and asked to extract launch speed and landing time. Marker traps focus on candidates who plug the resultant speed in place of a component or forget to use $r = R_E + h$ on the vertical axis when a height above the launch is involved.

IA2 student experiment. The standard projectile IA2 measures range as a function of launch angle for a fixed launch speed (small projectile launcher or a ball-bearing on a ramp). A strong report linearises by plotting $R$ against $\sin(2\theta)$ (slope $= v_0^2 / g$ ), reports a $v_0$ that agrees with a direct measurement, and discusses air-resistance and release-height systematic effects in the evaluation. The IA2 criteria reward the design justification, the data, and the evaluation in that order.

Common traps

Mixing up horizontal and vertical equations: Horizontal velocity never changes (with air resistance neglected). Vertical velocity changes by $-9.8$ m/s every second. Set up two separate columns of working and never let a $\cos$ leak into a vertical equation.
Forgetting the sign of $g$: If you take up as positive, $g$ enters the equations as $-9.8 \text{ m/s}^2$ . If you take down as positive, $g$ is $+9.8 \text{ m/s}^2$ . Pick a convention and stick to it for the whole question.
Using the speed instead of a component: $v_0 = 25$ m/s at $40°$ does not mean the horizontal velocity is $25$ m/s. You must resolve into components first.
Treating horizontally thrown objects as having $v_{0y} = v_0$: If a stone is thrown horizontally off a cliff, $v_{0y} = 0$ . The full speed is in the horizontal direction.
Using the level-ground range formula on uneven terrain: $R = v_0^2 \sin(2\theta) / g$ only works when launch height equals landing height. Otherwise solve the vertical equation for $t$ first, then compute $x = v_{0x} t$ .
Forgetting units: Markers deduct for missing units (m, s, m/s) even when the number is correct.

Examples in context

Example 1. A Bundaberg sugar mill conveyor discharges crushed bagasse horizontally at $4.0 \text{ m s}^{-1}$ from $3.0 \text{ m}$ height. Vertical time to fall is $t = \sqrt{2h/g} = 0.782 \text{ s}$ ; horizontal range is $x = v_x t = 3.13 \text{ m}$ . Engineers position the receiving hopper accordingly. The QCAA Unit 3 dot point decomposition into horizontal ( $v_x =$ constant) and vertical ( $a = -g$ ) is exactly this calculation.

Example 2. A Cairns hospital ballistic-trauma study models a $30 \text{ m s}^{-1}$ throw at $35^\circ$ above horizontal. Components: $v_x = 24.6 \text{ m s}^{-1}$ , $v_y = 17.2 \text{ m s}^{-1}$ . Time aloft $t = 2v_y/g = 3.51 \text{ s}$ , range $x = v_x t = 86.3 \text{ m}$ , peak height $h = v_y^2/(2g) = 15.1 \text{ m}$ . QCAA EA Unit 3 Paper 2 sets this template with one missing input.

Try this

Q1. State the assumption made when analysing projectile motion in QCAA Unit 3. [2 marks]

Cue. Constant downward $g$ , negligible air resistance.

Q2. A ball is thrown at $20 \text{ m s}^{-1}$ at $30^\circ$ above horizontal. Calculate the time aloft and the range, $g = 9.8 \text{ m s}^{-2}$ . [3 marks]

Cue. $v_y = 10 \text{ m s}^{-1}$ , $t = 2v_y/g = 2.04 \text{ s}$ ; $x = v_x t = 17.3 \times 2.04 = 35.3 \text{ m}$ .

Q3. A bagasse conveyor discharges horizontally at $4.0 \text{ m s}^{-1}$ from $3.0 \text{ m}$ above the hopper. (a) Calculate the time to fall and the range. (b) Determine the speed and angle of impact below horizontal. (c) Identify one engineering assumption that limits accuracy. [3+3+2 marks; ISMG: Analysis and interpretation, Evaluation]

Cue. (a) $t = 0.782 \text{ s}$ , $x = 3.13 \text{ m}$ ; (b) $v_y = 7.67 \text{ m s}^{-1}$ , speed $\sqrt{16+58.8} = 8.65 \text{ m s}^{-1}$ , angle $62^\circ$ ; (c) ignoring drag on the irregular bagasse.

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice5 marks

A javelin is released from a height of 1.8 m above the ground at 22 m/s at 38 degrees above the horizontal. Ignoring air resistance and using g = 9.8 m/s^2, calculate (a) the maximum height reached above the ground, (b) the time taken from release until the javelin lands, and (c) the horizontal range from release to landing.

Show worked answer →

Resolve the launch velocity.

$v_{0x} = 22 \cos 38° = 17.34$ m/s.
$v_{0y} = 22 \sin 38° = 13.54$ m/s.

(a) Maximum height above the ground. Above the release point, use $v_y^2 = v_{0y}^2 - 2 g \Delta y$ with $v_y = 0$ :

$\Delta y = \frac{v_{0y}^2}{2g} = \frac{13.54^2}{2 \times 9.8} = 9.35$ m.

Maximum height above the ground = 1.8 + 9.35 = 11.2 m.

(b) Time of flight. Take up as positive and let the displacement at landing be $-1.8$ m (the ground is below the release point).

$-1.8 = 13.54 t - \tfrac{1}{2} \times 9.8 \times t^2$
$4.9 t^2 - 13.54 t - 1.8 = 0$
$t = \frac{13.54 + \sqrt{13.54^2 + 4 \times 4.9 \times 1.8}}{2 \times 4.9} = \frac{13.54 + 14.79}{9.8} = 2.89$ s.

(c) Horizontal range.

$R = v_{0x} t = 17.34 \times 2.89 = 50.1$ m.

Markers reward the explicit resolution of components, the sign convention shown for $y$ , use of the quadratic for the off-level landing, and answers with units and 2 to 3 significant figures.

Practice3 marks

A small stone is thrown horizontally at 15 m/s from a 30 m cliff. Determine the time taken to reach the water below and the horizontal distance from the base of the cliff. Use g = 9.8 m/s^2.

Show worked answer →

Horizontal and vertical motions are independent. Initial vertical velocity is zero because the throw is horizontal.

Time of flight. Using $y = \tfrac{1}{2} g t^2$ with $y = 30$ m:

$t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{60}{9.8}} = 2.47$ s.

Horizontal distance. Horizontal velocity is constant.

$x = v_x t = 15 \times 2.47 = 37.1$ m.

Markers reward an explicit statement that $v_{0y} = 0$ , the identification of gravity as the only force, and answers with units. Stating that air resistance is neglected secures the assumption mark.

Practise this

Sources & how we know this

Physics General senior syllabus 2025 — QCAA