Projectile motion in two dimensions: VCE Physics Unit 3

Syllabus dot point

“Investigate and analyse theoretically and practically the motion of projectiles near Earth's surface including a qualitative description of the effects of air resistance”

9 min readUpdated 21 May 2026

Quick answer

Projectile motion is solved by splitting the launch velocity into horizontal and vertical components and applying constant-velocity equations horizontally and SUVAT vertically with $g = 9.8$ m/s squared, linked by the shared time of flight, with air resistance qualitatively reducing both range and maximum height.

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What this dot point is asking
The answer
Examples in context
Try this
Exam-style questions

What this dot point is asking

VCAA wants you to model a projectile (an object moving only under gravity) by splitting its velocity into independent horizontal and vertical components, then applying the equations of motion to each axis. You also need to describe qualitatively how air resistance changes the trajectory.

The answer

A projectile is any object in flight that is subject only to gravity (we ignore air resistance for the calculations). The trick is that 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 ignored), so horizontal velocity is constant.

x = v_{0x} t

Vertical motion

The only acceleration is gravity, $a_y = -g = -9.8$ m/s squared (taking up as positive). Use SUVAT:

v_y = v_{0y} - gt

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

v_y^2 = v_{0y}^2 - 2gy

Key features of the trajectory

The path is a parabola. At maximum height $v_y = 0$ , so $h_{\max} = \frac{v_{0y}^2}{2g}$ .

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

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

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

Air resistance (qualitative)

Air resistance (drag) acts opposite to the velocity vector at every instant and increases with speed (roughly $F_{drag} \propto v^2$ ). Its effects on a projectile:

Range falls. Horizontal velocity decreases throughout the flight rather than staying constant.
Maximum height falls. Drag removes kinetic energy on the way up.
Trajectory loses symmetry. The descent is steeper than the ascent because the ball has less horizontal velocity by the time it falls.
Optimum launch angle is below 45 degrees for maximum range with drag (because the ball benefits from spending less time in the air).

VCAA expects qualitative description only. No numerical drag questions.

Worked example

A ball is kicked from ground level at $v_0 = 20$ m/s at $\theta = 40°$ above horizontal. Find the maximum height, time of flight, and range.

Resolve: $v_{0x} = 20 \cos 40° = 15.32$ m/s; $v_{0y} = 20 \sin 40° = 12.86$ m/s.

Maximum height: $h_{\max} = \frac{v_{0y}^2}{2g} = \frac{12.86^2}{19.6} = 8.44$ m.

Time of flight: $t = \frac{2 v_{0y}}{g} = \frac{2 \times 12.86}{9.8} = 2.62$ s.

Range: $R = v_{0x} t = 15.32 \times 2.62 = 40.2$ m.

Try it: Projectile motion calculator - enter launch speed, angle and height and get the range, max height and trajectory.

Common traps

Mixing horizontal and vertical equations: Horizontal velocity is constant (no drag). Vertical velocity changes by $9.8$ m/s each second. Two separate columns of working.
Forgetting the sign of $g$: Pick a positive direction (up or down) and apply it consistently to $v_{0y}$ , $a_y$ and $y$ .
Using the launch speed instead of a component: $v_0 = 25$ m/s at $40°$ does not mean horizontal velocity is $25$ m/s. Resolve first.
Treating a horizontally thrown object as having $v_{0y} = v_0$: If a stone is thrown horizontally off a cliff, $v_{0y} = 0$ .
Treating air resistance as a constant force: Drag depends on speed; expect a qualitative description, not a numerical one.

Examples in context

Example 1. MCG cricket ball six-hit trajectory. A batter at the MCG hits a $156$ g cricket ball at $35$ m s $^{-1}$ at $35^\circ$ above horizontal. Resolving: $v_x = 35 \cos 35^\circ = 28.7$ m s $^{-1}$ ; $v_y = 35 \sin 35^\circ = 20.1$ m s $^{-1}$ . Time to peak: $20.1/9.8 = 2.05$ s. Maximum height: $v_y^2/(2g) = 404/19.6 = 20.6$ m, well below the MCG light tower. Total time of flight (return to ground): $2 \times 2.05 = 4.10$ s. Horizontal range: $28.7 \times 4.10 = 117.7$ m, clear of the $86$ m square boundary, scoring a six. Air drag reduces actual range by approximately $10$ - $15$ %.

Example 2. Tullamarine fire-bomber water drop. A NAFC Air-Tractor AT-802F fire-bombing aircraft, flying at $50$ m s $^{-1}$ at $60$ m altitude over a Wombat State Forest bushfire, releases water tanks. Treating each release as a projectile with $v_x = 50$ m s $^{-1}$ and $v_y = 0$ initially: time to ground $t = \sqrt{2h/g} = \sqrt{120/9.8} = 3.50$ s. Forward travel during fall: $50 \times 3.50 = 175$ m. So the pilot must release the load $175$ m before the target. Impact speed: $\sqrt{50^2 + (gt)^2} = \sqrt{2500 + 1177} = 60.6$ m s $^{-1}$ . Air resistance disperses the water column over a $30$ m drop pattern.

Try this

Q1. Explain why horizontal and vertical motions of a projectile can be treated separately. [2 marks]

Cue. Gravity acts only vertically; no horizontal acceleration in absence of air resistance. Time variable links them.

Q2. A ball is projected horizontally at $15$ m s $^{-1}$ from a $20$ m cliff. Calculate (a) the time to ground, and (b) the horizontal range. [4 marks]

Cue. (a) $t = \sqrt{40/9.8} = 2.02$ s. (b) $R = 15 \times 2.02 = 30.3$ m.

Q3. Refer to the MCG cricket six. (a) Calculate maximum height for $u = 35$ m s $^{-1}$ at $35^\circ$ . (b) Determine total flight time. (c) Outline qualitatively how air resistance would alter the trajectory. [2+2+2 marks]

Cue. (a) $20.6$ m. (b) $4.10$ s. (c) Reduces both maximum height and horizontal range; trajectory becomes asymmetric with steeper descent.

Exam-style questions

Questions in the style of VCAA 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.

Practice4 marks

A stone is thrown horizontally at 15 m/s from the top of a 20 m cliff. Calculate the time taken to reach the ground and the horizontal distance from the base of the cliff at which the stone lands. (Use g = 9.8 m/s^2 and ignore air resistance.)

Show worked answer →

Horizontal and vertical motions are independent. Take down as positive.

Initial vertical velocity $v_{0y} = 0$ (thrown horizontally).

Time to land. Using $y = \frac{1}{2} g t^2$ with $y = 20$ m:

$t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{2 \times 20}{9.8}} = 2.02$ s.

Horizontal distance. Horizontal velocity is constant at $15$ m/s.

$x = v_x t = 15 \times 2.02 = 30.3$ m.

Markers reward the explicit statement that $v_{0y} = 0$ , the correct use of $g = 9.8$ m/s squared, and final answers with units to two or three significant figures.

Practice5 marks

A ball is launched from ground level at 24 m/s at an angle of 35 degrees above horizontal. Calculate the maximum height, time of flight and horizontal range. Then explain qualitatively how each answer would change if air resistance were not negligible.