Projectile Motion¶
Syllabus mapping
PY-12-01 — Advanced Mechanics. Covers: analysing projectile motion by resolving into independent horizontal and vertical components; modelling and solving for the relationships between launch angle, initial velocity, launch height, maximum height, time of flight, final velocity and range.
Wording above is paraphrased for teaching use — cross-check against the official syllabus PDF (see Resources) before treating any line as verbatim NESA text.
You need to know: how to split a launch velocity into horizontal and vertical components, then treat those two directions completely independently — one has zero acceleration, the other has constant acceleration \(g\) — and combine them to solve for time of flight, maximum height and range.
Assumed knowledge check¶
Before this page makes sense, you should be comfortable with:
- Quantities of Motion — resolving a vector into components, \(A_x = A\cos\theta\), \(A_y = A\sin\theta\)
- Motion Relationships — the SUVAT equations, applied here separately to each axis
- Trigonometry — SOH-CAH-TOA
Core¶
The key idea: two independent 1D problems¶
A projectile (anything launched and then left to move under gravity alone — no engine, no air resistance) undergoes two motions at the same time, completely independent of each other:
- Horizontal: zero acceleration. Horizontal velocity is constant throughout the flight.
- Vertical: constant acceleration \(g\) downward, exactly like an object dropped or thrown straight up.
The only thing linking the two directions is time — both start at \(t=0\) and both are evaluated at the same \(t\) when you want the projectile's position or velocity. Splitting the vector like this turns one hard 2D problem into two easy 1D SUVAT problems.
For a launch velocity \(v\) at angle \(\theta\) above the horizontal:
\(v_x\) stays fixed for the whole flight. \(v_y\) changes under \(a=-g\) exactly like Year 11's vertical SUVAT problems — it decreases to zero at the peak, then becomes increasingly negative (downward) on the way down.
Sign convention
Pick "up" as positive before you start (matching the sign-convention habit from Motion Relationships), and use \(a=-9.8\ \text{m s}^{-2}\) throughout. Mixing signs mid-problem is the single most common source of projectile motion errors.
Launched and landing at the same height¶
This is the simplest case — level ground, launch and landing heights equal.
Time of flight, from symmetry (time up = time down): the projectile returns to \(v_y=0\)'s mirror image, so
Maximum height, using \(v^2=u^2+2as\) on the vertical axis with final vertical velocity zero at the peak:
Range (horizontal distance travelled), since horizontal velocity is constant:
Worked example — level launch
A ball is launched at \(20\ \text{m s}^{-1}\) at \(40^\circ\) above the horizontal, from and to the same height. Find the time of flight, maximum height, and range.
Resolve the launch velocity:
Time of flight:
Maximum height:
Range:
Launched from a height¶
When the launch point is higher than the landing point (fired horizontally off a cliff, rolled off a table, thrown from a window), the up-down symmetry trick above no longer applies — you can't use \(t=2v_y/g\), because the projectile falls further than it rose. Go back to first principles: use \(s=ut+\tfrac{1}{2}at^2\) on the vertical axis with the actual vertical displacement (usually negative, since it ends up below where it started), then use that \(t\) for the horizontal distance.
Worked example — launched from a height
A ball rolls off a table \(1.2\ \text{m}\) high with a horizontal speed of \(3\ \text{m s}^{-1}\). Find the time to land and the horizontal distance travelled.
Vertical axis: \(u_y=0\) (purely horizontal launch), \(s=-1.2\ \text{m}\), \(a=-9.8\ \text{m s}^{-2}\):
Horizontal axis: \(v_x\) never changes.
Notice the horizontal speed had no effect at all on how long the ball was in the air — that's decided entirely by the vertical drop. This surprises most students the first time; it's worth sitting with.
Final velocity¶
At any point in the flight, the projectile's instantaneous velocity is the vector sum of its (still-constant) \(v_x\) and its (SUVAT-updated) \(v_y\) at that moment — resolve back to magnitude/direction with Pythagoras and \(\tan^{-1}\), exactly as in Quantities of Motion.
🖼️ Diagram needed: a trajectory (parabolic path) with velocity vectors drawn at launch, peak, and landing, each split into its \(v_x\)/\(v_y\) components — should visually show \(v_x\) staying the same length throughout while \(v_y\) shrinks to zero then grows again in the opposite direction. Static, Excalidraw-style, matching the free-body diagram colour convention once that's settled in Forces and Motion. TODO — placeholder until sourced/drawn.
Aboriginal and Torres Strait Islander content — placeholder, not yet written
NESA lists the use of projectile motion by Aboriginal and/or Torres Strait Islander Peoples (e.g. woomera-launched spears, boomerangs) as explicit, examinable syllabus content — not an optional cultural add-on (see project.md Section 0). This isn't something to paraphrase generically or guess at: it needs to be researched and written properly, ideally with input from your school's Aboriginal Education team, a relevant community-verified source, or NESA's own syllabus support materials, so it's accurate and respectfully presented rather than a token line.
Deliberately left unwritten here rather than filled with unverified content — flagging clearly so it doesn't get missed, per the project's own checklist (project.md Section 7, "don't skip as 'extra'").
Practical¶
Suggested prac: projectile motion investigation — launch a projectile (spring launcher, ramp and ball, or similar) at a fixed speed across a range of angles, measuring range and/or time of flight, and compare against the predicted range formula below. TODO — link this school's actual prac instructions once written.
Advanced
The range formula, and why 45° is special. Combining \(t=2v_y/g\) and \(R=v_xt\) with \(v_x=v\cos\theta\), \(v_y=v\sin\theta\):
(using the identity \(2\sin\theta\cos\theta=\sin2\theta\)). Since \(\sin(2\theta)\) is maximised when \(2\theta=90^\circ\), the range is maximum at launch angle \(45^\circ\) — and, because \(\sin(2\theta)=\sin(180^\circ-2\theta)\), any two launch angles that add to \(90^\circ\) (e.g. \(30^\circ\) and \(60^\circ\)) give the same range, just with different times of flight and maximum heights. Worth checking with real numbers if it doesn't feel obvious yet.
Multi-stage problems. Some questions combine a projectile phase with something else — e.g. a ball launched from a moving platform, or a projectile that bounces and relaunches. Treat each phase separately: finish the projectile-motion analysis for one phase completely (final position, final velocity) before using that as the starting condition for the next phase.
Extension
Beyond the Physics 11–12 syllabus — won't appear in the HSC, included for interest / depth study inspiration.
📎 Depth study idea
Real projectiles experience air resistance, which is not constant — it grows with speed (often roughly as \(v^2\)), so the horizontal motion is no longer unaccelerated and the vertical motion no longer follows clean SUVAT. The resulting trajectory is shorter-ranged and asymmetric (steeper on the way down than the way up) compared to the ideal parabola covered above. This has no clean algebraic solution — it requires numerical methods (small time-step simulation) to model properly, which makes it a genuinely good depth study: pick a real projectile (a thrown ball, a dropped object with a parachute, a model rocket) and compare the ideal parabolic prediction against a simple simulated or measured trajectory, discussing where and why they diverge.
Video/visual resources¶
- 🖥️ PhET Simulation — Projectile Motion — TODO: confirm current PhET link
- 🎥 Khan Academy — TODO: source a 2D projectile motion explainer. Must clearly show the horizontal/vertical independence, ideally with a launched-from-height example as well as level ground. Essential.
- 🎥 Physics High — TODO: check for a NSW-syllabus-aligned projectile motion video, ideally covering the range formula and the 45°-optimum result specifically, since that's a common exam extension question. Nice-to-have alongside the Khan Academy one above.
Check yourself¶
-
A stone is thrown horizontally at \(8\ \text{m s}^{-1}\) from a cliff \(20\ \text{m}\) high. Find (a) the time to reach the ground and (b) the horizontal distance travelled.
Answer
(a) Vertical axis: \(u_y=0\), \(s=-20\ \text{m}\), \(a=-9.8\ \text{m s}^{-2}\):
\(-20 = -\tfrac{1}{2}(9.8)t^2 \implies t = \sqrt{40/9.8} \approx 2.02\ \text{s}\)
(b) \(x = v_xt = 8 \times 2.02 \approx 16.2\ \text{m}\)
-
A ball is launched at \(15\ \text{m s}^{-1}\) at \(50^\circ\) above the horizontal from level ground. Find the maximum height and the range.
Answer
\(v_x = 15\cos50^\circ \approx 9.64\ \text{m s}^{-1}\), \(v_y = 15\sin50^\circ \approx 11.5\ \text{m s}^{-1}\)
\(H = \dfrac{v_y^2}{2g} = \dfrac{11.5^2}{2\times9.8} \approx 6.75\ \text{m}\)
\(t = \dfrac{2v_y}{g} = \dfrac{2\times11.5}{9.8} \approx 2.35\ \text{s}\), so \(R = v_xt = 9.64 \times 2.35 \approx 22.7\ \text{m}\)
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Two projectiles are launched from level ground at the same speed, one at \(20^\circ\) and one at \(70^\circ\) above the horizontal. Explain, without calculating, why they land the same distance away.
Answer
\(20^\circ\) and \(70^\circ\) add to \(90^\circ\), and the range formula \(R=\dfrac{v^2\sin(2\theta)}{g}\) gives the same value for any pair of angles that sum to \(90^\circ\) (since \(\sin(2\times20^\circ)=\sin40^\circ\) and \(\sin(2\times70^\circ)=\sin140^\circ=\sin40^\circ\) as well). They won't have the same time of flight or maximum height, though — the steeper angle spends longer in the air and goes higher.