A projectile is anything moving freely under gravity in two dimensions — a thrown ball, a launched cannonball, a diver leaving the board. The big idea is beautifully simple: split the motion into a horizontal part and a vertical part, and treat them completely separately. Each part is just a SUVAT problem.
📘 What you need to know
A projectile moves freely under gravity in a 2D plane (no engine, air resistance ignored).
Horizontal and vertical motion are independent — solve each one separately with SUVAT.
Horizontal: velocity is constant (a = 0), because nothing pushes it sideways.
Vertical: acceleration is g = 9.81 m s⁻² (gravity), always pointing down.
Split a launch velocity u at angle θ into: horizontal = u cos θ, vertical = u sin θ.
At maximum height the vertical velocity is zero (but horizontal velocity carries on).
Time, t, is the shared link between the horizontal and vertical motions.
The big idea: two motions in one
The single most important thing about projectiles is that the horizontal and vertical motions don’t affect each other. Gravity only acts downward, so it only changes the vertical velocity. Sideways, nothing pushes the object, so its horizontal velocity just stays the same the whole flight.
The horizontal velocity (blue) never changes; the vertical velocity (amber) shrinks to zero at the top, then grows on the way down. Together they trace a parabola.
Splitting the launch velocity
When something is launched at a speed u at an angle θ above the horizontal, your first job is almost always to resolve that velocity into its two components using trigonometry. Draw the velocity as the hypotenuse of a right-angled triangle:
Resolve the launch velocity: the horizontal component is u cos θ, the vertical component is u sin θ.
Resolving the launch velocity
horizontal: uH = u cos θ • vertical: uV = u sin θ
Horizontal vs vertical — the master table
Keep these two columns separate in every projectile problem. The only thing they share is the time.
Quantity
Horizontal
Vertical
Velocity
constant (u cos θ)
changes; zero at the top
Acceleration
0
g = 9.81 m s⁻² (downward)
Use it to find
range (s = ut)
max height, time of flight
Key terms:time of flight = total time in the air; maximum height = where vertical velocity = 0; range = horizontal distance travelled.
When air resistance is ignored, the path is perfectly symmetrical. The time to reach the top is exactly half the total flight time — so you can find one half and double it.
🧭 Recipe — solving a projectile problem
Resolve the launch velocity into u cos θ (horizontal) and u sin θ (vertical).
Set up two SUVAT columns, one horizontal (a = 0) and one vertical (a = g). Choose a positive direction and stick to it.
Use the vertical motion to find the time (e.g. v = u + at, with v = 0 at the top).
Feed that time into the horizontal motion (s = ut) to get the range.
Worked examples
WE 1
Maximum height of a thrown ball
A ball is thrown from point P at 12 m s⁻¹ at 50° above the horizontal. Ignoring air resistance, find the maximum height it reaches.
Only vertical motion matters for heightu = 12 sin 50° , v = 0 (at the top), a = −9.81Use v² = u² + 2as → s = (v² − u²) ÷ 2as = (0 − (12 sin 50°)²) ÷ (2 × −9.81)s = −(9.19²) ÷ (−19.62)max height ≈ 4.3 m
WE 2
Range of a stone thrown off a cliff
A stone leaves the top of a 50.0 m cliff at 30.0 m s⁻¹, directed 25.0° below the horizontal. It hits the ground with a vertical velocity of 33.8 m s⁻¹. Find the horizontal distance D from the base of the cliff.
Resolve the launch velocityu_vert = 30 sin 25° = 12.68 m s⁻¹u_horiz = 30 cos 25° = 27.19 m s⁻¹Vertical: find time with v = u + at33.8 = 12.68 + 9.81tt = (33.8 − 12.68) ÷ 9.81 = 2.15 sHorizontal: D = u_horiz × tD = 27.19 × 2.15D ≈ 58 m
WE 3
A horizontally-launched stunt rider
A stunt rider leaves a ramp moving horizontally from 1.25 m above the ground and lands 10 m away. Ignoring air resistance, find the take-off speed. (Take g = 9.81 m s⁻².)
Vertical: find the fall time (u_vert = 0)Use s = ½at² → 1.25 = ½ × 9.81 × t²t² = (2 × 1.25) ÷ 9.81 = 0.2548t = √0.2548 = 0.505 sHorizontal: speed = distance ÷ timeu = 10 ÷ 0.505take-off speed ≈ 19.8 m s⁻¹
💡 Top tips
Always split into horizontal and vertical first. Never mix a vertical distance with a horizontal velocity.
At the highest point, vertical velocity = 0 (but horizontal velocity carries on unchanged).
Time is the bridge — find it from the vertical motion, then use it in the horizontal motion.
Use symmetry: with no air resistance, time up = time down, so total time = 2 × time to the top.
Be careful with sin and cos: vertical uses sin θ, horizontal uses cos θ.
⚠ Common mistakes
Mixing horizontal and vertical quantities — keep them in separate columns at all times.
Swapping sin θ and cos θ — vertical component is u sin θ, horizontal is u cos θ.
Sign errors — pick one positive direction (often “up”) and apply it consistently to every vector.
Thinking gravity affects the horizontal motion — it doesn’t; horizontal velocity is constant.
Forgetting unit conversions (cm, km → m) before substituting.
Up next: Fluid Resistance — what changes once we stop ignoring air resistance, and how a real falling object behaves differently from an ideal projectile.
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