Do you include air resistance in HSC projectile motion?

No. The HSC model assumes no air resistance and a constant downward acceleration of 9.8 m s^−2. You may be asked to describe, in words, how air resistance would change a real path (shorter range, lower peak, steeper landing), but calculations ignore it.

What value of g should I use in the HSC?

Use 9.8 m s^−2, the value on the HSC data sheet, unless the question gives a different one (for example, on the Moon).

Which projectile equations are on the HSC data sheet?

The 2019 data sheet gives the general equations of motion: s = ut + ½at^2, v = u + at and v^2 = u^2 + 2as. You apply them separately to the horizontal and vertical directions. The component formulas u cos θ and u sin θ are not on the sheet.

Why is the horizontal velocity of a projectile constant?

Because there is no horizontal force once it is launched (with no air resistance, only gravity acts, and gravity is vertical). No net horizontal force means no horizontal acceleration.

What angle gives the maximum range?

45°, but only when the projectile lands at the same height it was launched from. Launched from above the landing point, the best angle is a little less than 45°.

A ball is kicked at 40° above the horizontal on level ground. At the highest point of its path, which statement is correct?

Its velocity is horizontal and its acceleration is 9.8 m s^−2 down.

Its velocity is zero and its acceleration is zero.

Its velocity is zero and its acceleration is 9.8 m s^−2 down.

Its velocity is horizontal and its acceleration is zero.

HSC Physics · Module 5 · Advanced Mechanics · 10 min read

Projectile motion

A projectile is anything that flies through the air with only gravity acting on it: a kicked ball, a stone off a cliff, a package dropped from a drone. In HSC Physics every projectile question comes down to one idea: the horizontal and vertical motions are separate, and time is the only thing they share.

This guide gives you that idea in a few minutes, the equations to use, the mistakes that cost marks, a quick quiz and a full worked example. Like the HSC course, it ignores air resistance throughout.

What you need to know

Treat a projectile's horizontal and vertical motion separately, assuming gravity gives a constant downward acceleration and there is no air resistance.
From that model, work out how these are related: launch velocity, launch angle, greatest height, total flight time, landing velocity, launch height and horizontal range.
Test the projectile equations with your own data in a practical investigation (for example, measuring the range of a ball rolled off a bench at different speeds).
Use the straight-line equations of motion (constant velocity horizontally, uniform acceleration vertically) to solve problems and make predictions.

The equations (from the 2019 HSC data sheet unless marked):

s = ut + ½at²Use it twice: horizontally (a = 0, so s_x = u_xt) and vertically (a = −9.8 m s⁻² if up is positive).

v = u + atVertical velocity at any time. Horizontally v_x = u_x for the whole flight.

v² = u² + 2asVertical only: the greatest height (v_y = 0) or the vertical speed at landing.

u_x = u cos θ u_y = u sin θ not on the data sheetResolve the launch velocity first (θ measured from the horizontal).

Dot points: PH-M5-01, PH-M5-02, PH-M5-03, PH-M5-04 (2017 Physics syllabus, HSC 2026 and 2027), in our own words.

Mini lesson

Two motions at once

Once a projectile is launched, the only force on it is its weight, which points straight down. There is no horizontal force, so there is no horizontal acceleration: the horizontal velocity stays the same for the whole flight.

Vertically, the projectile accelerates downward at g = 9.8 m s⁻². Its vertical velocity drops by 9.8 m s⁻¹ every second on the way up, is zero at the top, and grows by 9.8 m s⁻¹ every second on the way down.

Blue: the horizontal velocity is the same at every point. Orange: the vertical velocity shrinks to zero at the top, then grows downward. The acceleration is 9.8 m s⁻² down the whole time, even at the top.

Step one, always: resolve the launch velocity

The equations work on one direction at a time, so split the launch velocity u into its components before anything else. With θ measured from the horizontal, the horizontal part is u cos θ and the vertical part is u sin θ.

The side next to θ uses cos; the side opposite θ uses sin. Check: each component must be smaller than u.

Time links the two directions

Horizontal and vertical motion have different equations but the same clock. The usual plan is: use the vertical motion to find the time of flight, then use that time horizontally to find the range (s_x = u_xt).

This is why a ball rolled off a table and a ball dropped from the same height hit the floor together: their vertical motions are identical. The rolled ball just travels sideways at the same time.

Greatest height: vertical only, v_y = 0, so h = u_y² ÷ 2g above the launch point.
Time to the top: v_y = 0 = u_y − gt, so t = u_y ÷ g.
Level ground only: the flight is symmetric, so total time = 2u_y ÷ g and the landing speed equals the launch speed.
Launch and landing at different heights: the flight is not symmetric. Put the real vertical displacement into s = ut + ½at² and solve the quadratic.

Reading the motion from graphs

HSC questions often give graphs instead of numbers. A horizontal-velocity graph is a flat line. A vertical-velocity graph is a straight line with gradient −9.8 m s⁻², crossing zero at the top of the flight. The area under each velocity graph is that direction's displacement.

Horizontal velocity is constant; vertical velocity changes at a steady −9.8 m s⁻² (up taken as positive).

Angle and range on level ground

For a launch and landing at the same height, the range is greatest at 45°, and angles that add to 90° (such as 30° and 60°) give the same range at the same launch speed. The higher angle stays in the air longer and goes higher, but moves sideways more slowly.

Range depends on the square of the launch speed: double the speed at the same angle and the range is four times as far. None of this depends on mass, because every object falls with the same g.

Common mistakes

Putting the full launch speed u straight into a vertical or horizontal equation.Resolve first: u_x = u cos θ and u_y = u sin θ, then use each in its own direction.
Saying the velocity (or the acceleration) is zero at the top of the flight.Only the vertical velocity is zero at the top. The projectile is still moving at u_x, and the acceleration is 9.8 m s⁻² down the whole time.
Using 'time to the top × 2' when the projectile lands higher or lower than it was launched.That shortcut is for level ground only. Use the real vertical displacement in s = ut + ½at².
Mixing signs: u_y positive, but g also positive, so the ball never comes down.Write your sign convention down (up = +), then a = −9.8 m s⁻² and any landing point below the launch has a negative s.
Thinking a faster horizontal launch makes an object take longer to fall.The fall time depends only on the vertical motion. Horizontal speed changes how far it goes, not how long it takes.
Rounding the time to one significant figure, then using it for the range.Keep the full calculator value for the time and round only the final answer.

Worked example

A basketball is released 2.10 m above the floor with a speed of 9.0 m s⁻¹ at 50° above the horizontal. It passes through the hoop, 3.05 m above the floor, while it is coming down.

Calculate the horizontal distance from the release point to the hoop. Ignore air resistance.

(4 marks)

I'll think out loud, the way you should in the exam, using KNOW-UD: Knowns, Not known, Organise, Work it out, Units, Direction.

Prediction first. On level ground this launch would travel u² sin 2θ ÷ g = 81 × sin 100° ÷ 9.8 ≈ 8.1 m. The hoop is about a metre higher than the release, so the ball reaches it a little before it would get back to release height. I predict a bit under 8 m, somewhere around 7 m.

K · Knowns

I sketch it first, with every value on the sketch, then list them as symbols. The key number is the vertical displacement: the hoop is above the release point, so it's positive.

u = 9.0 m s⁻¹, θ = 50°
Δy = 3.05 − 2.10 = +0.95 m (up is +)
a_y = −9.8 m s⁻², a_x = 0

Knowns on the sketch: u = 9.0 m s⁻¹ at 50°, release 2.10 m, hoop 3.05 m, so Δy = +0.95 m. Unknown: the horizontal distance x.

N · Not known

The question wants a horizontal distance, so the last line has to be x in metres. To get x I need the time, and the time comes from the vertical motion.

x = ? m (and on the way: t = ? s)

O · Organise

Up is positive. I resolve the launch velocity, then write the vertical equation in symbols before any numbers. The ball passes 0.95 m above the release point twice, once rising and once falling, so I expect two answers for t and I'll need the later one.

u_x = u cos θ, u_y = u sin θ
Δy = u_yt + ½a_yt²
x = u_xt

W · Work it out

Components first, keeping the full calculator values. Then the quadratic. I'll write it in standard form so the substitution line is clear.

u_x = 9.0 cos 50° = 5.785 m s⁻¹, u_y = 9.0 sin 50° = 6.894 m s⁻¹
0.95 = 6.894t − 4.9t² → 4.9t² − 6.894t + 0.95 = 0
t = [6.894 ± √(6.894² − 4 × 4.9 × 0.95)] ÷ (2 × 4.9) = (6.894 ± 5.377) ÷ 9.8
t = 0.155 s (rising) or t = 1.252 s (falling): the ball is coming down, so t = 1.252 s
x = u_xt = 5.785 × 1.252 = 7.24 m

U · Units

Metres per second times seconds gives metres, which is right for a distance. The data are given to two or three significant figures, so I'll give two.

x ≈ 7.2 m

D · Direction

It's a horizontal distance, so I say which way: from the release point towards the hoop. As a sense check on the 'coming down' choice: at 1.25 s the vertical velocity is 6.894 − 9.8 × 1.252 = −5.4 m s⁻¹, negative, so it is falling. Good.

x = 7.2 m horizontally, from the release point towards the hoop

The hoop is about 7.2 m horizontally from the release point.

Check against the prediction. I predicted a bit under 8 m, around 7 m, because the hoop is above the release height. 7.2 m is just under the 8.1 m level-ground range, so it fits.

HSC advice

What markers keep saying

Resolve the launch velocity before using any equation, and check you have cos and sin the right way round. Using the full launch speed in the vertical equations, or swapping sin and cos, were the slips to avoid.NESA marking feedback, 2025, summarised
Don't assume the flight is symmetric when the launch and landing heights differ. Strong answers used the actual vertical displacement and solved for the time, rather than doubling the time to the top.NESA marking feedback, 2022, summarised
Pick a sign convention for the vertical direction, state it, and keep it. Give the vertical velocity and acceleration the right signs for your convention (up = +, so a = −9.8 m s⁻²) in every equation.NESA marking feedback, 2021, summarised
When a graph or table gives the motion, read the initial velocity carefully: an average velocity worked out from the data is not the launch velocity unless there is no acceleration.NESA marking feedback, 2020, summarised
Show the formula, the substitution and the answer with a unit. Correct working earns marks even when the final number slips; a bare answer earns none if it's wrong.General advice from NESA marking feedback, 2021 to 2025, summarised

Summaries in our own words. Read NESA's own feedback with the exam packs: 2020 · 2021 · 2022 · 2025.

How often it is examined

Projectile motion questions were worth 46 marks across the seven HSC Physics exams from 2019 to 2025 (about 6.6 marks a paper), and appeared in 7 of 7 papers.

Marks per paper (Module 5, Projectile Motion). Source: CortexHSC's own tagging of the NESA HSC Physics exams 2019–2025, each question counted under its main topic, so treat the numbers as close, not exact.

FAQ

Do you include air resistance in HSC projectile motion?: No. The HSC model assumes no air resistance and a constant downward acceleration of 9.8 m s⁻². You may be asked to describe, in words, how air resistance would change a real path (shorter range, lower peak, steeper landing), but calculations ignore it.
What value of g should I use in the HSC?: Use 9.8 m s⁻², the value on the HSC data sheet, unless the question gives a different one (for example, on the Moon).
Which projectile equations are on the HSC data sheet?: The 2019 data sheet gives the general equations of motion: s = ut + ½at², v = u + at and v² = u² + 2as. You apply them separately to the horizontal and vertical directions. The component formulas u cos θ and u sin θ are not on the sheet.
Why is the horizontal velocity of a projectile constant?: Because there is no horizontal force once it is launched (with no air resistance, only gravity acts, and gravity is vertical). No net horizontal force means no horizontal acceleration.
What angle gives the maximum range?: 45°, but only when the projectile lands at the same height it was launched from. Launched from above the landing point, the best angle is a little less than 45°.

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