A projectile is anything thrown, kicked or launched that then flies through the air with only gravity acting on it — a football, a cannonball, a long-jumper. It looks complicated because the object moves sideways and up-and-down at the same time. But here’s the big secret that makes it easy: those two motions are completely independent. Split them apart, deal with each on its own, and a scary 2D problem becomes two simple 1D ones.
📘 What you need to know
A projectile moves freely under gravity only — we ignore air resistance
The horizontal and vertical motions are independent — solve them separately
Horizontally: velocity is constant (no force), so acceleration = 0
Vertically: acceleration = g = 9.81 m s−2, exactly like free fall
Resolve the launch velocity: horizontal = ucosθ, vertical = usinθ
At the very top, the vertical velocity is zero — a key fact for finding max height
What makes something a projectile?
A projectile is a particle moving freely — non-powered — under gravity, in a two-dimensional plane. Think of a ball thrown, a diver leaving a board, or a baseball off a bat. To keep the maths clean we make two assumptions:
Air resistance is negligible — the fluid the object moves through doesn’t slow it.
Gravity is constant — the object stays near the Earth’s surface, so g = 9.81 m s−2 throughout.
The big idea: split the motion in two
This is the whole topic in one sentence: the horizontal and vertical motions don’t affect each other. Gravity only pulls down, so it only changes the vertical motion. Sideways, nothing pushes or pulls, so the horizontal velocity just stays the same the entire flight.
horizontal
→ no force →
constant velocity (a = 0)
vertical
→ gravity →
free fall (a = g)
The launch velocity splits into a horizontal part (ucosθ, constant all flight) and a vertical part (usinθ, shrinking to zero at the top). Gravity acts down the whole time.
Splitting the launch velocity into components
An object launched at speed u and angle θ to the ground has that velocity pointing diagonally. To use the equations of motion, we break it into a flat (horizontal) part and an up (vertical) part using simple trigonometry — the classic SOH-CAH-TOA.
Resolve the launch velocity with trig: the horizontal part is ucosθ and the vertical part is usinθ.
Resolving the launch velocity
horizontal: ux = ucosθ • vertical: uy = usinθ
The mistake I see most often is students mixing up which component uses sine and which uses cosine. Here’s the anchor: the angle θ is measured from the horizontal ground. The side next to the angle (the horizontal one) is the “adjacent”, and adjacent goes with cos. The side opposite the angle (the vertical one) goes with sin. If you ever blank, sketch the triangle — the picture never lies.
The three things you’ll be asked to find
Almost every projectile question boils down to finding one of three quantities. Each comes from one of the two motions.
🎯 The three key quantities
Time of flight — how long it’s in the air. Comes from the vertical motion. Time to the top is half the total flight (for level ground).
Maximum height — the peak. At the top the vertical velocity is zero, so use v2 = u2 + 2as vertically.
Range — the horizontal distance. Comes from the horizontal motion: range = horizontal velocity × total time.
Quick recap: resolve u into ucosθ (flat) and usinθ (up). Do the vertical motion with a = g to get time and height; do the horizontal motion with a = 0 to get range. The two share only one thing: time.
WE 1
A ball is kicked from level ground at 25 m s−1, at 40° above the horizontal. Taking g = 9.81 m s−2, find (a) the maximum height, and (b) the total range.
Step 1 — resolve the launch velocity
horizontal: u cos40° = 25 × 0.766 = 19.15 m s⁻¹
vertical: u sin40° = 25 × 0.643 = 16.07 m s⁻¹
Part (a) — max height (vertical, v = 0 at top)
v² = u² + 2as → 0 = 16.07² − 2(9.81)H
H = 16.07² ÷ (2 × 9.81) = 258 ÷ 19.62max height = 13.2 mPart (b) — range (need total time first)
time up = 16.07 ÷ 9.81 = 1.64 s, so total T = 3.28 s
range = horizontal velocity × T = 19.15 × 3.28
range = 62.7 m
WE 2
A stunt rider drives horizontally off a ramp 1.25 m above the ground and lands 10 m away. Find the take-off speed. (g = 9.81 m s−2)
Step 1 — vertical motion gives the time
launched horizontally, so vertical u = 0
s = ut + ½at² → 1.25 = ½ × 9.81 × t²
t = √(2 × 1.25 ÷ 9.81) = √0.255 = 0.505 sStep 2 — horizontal motion gives the speed
horizontal velocity is constant: v = distance ÷ time
v = 10 ÷ 0.505take-off speed = 19.8 m s⁻¹The two motions meet only through the shared time of 0.505 s.
💡 Top tips
Always resolve first — split u into horizontal and vertical parts before doing anything else.
Time is the bridge — the vertical and horizontal motions share only the time of flight; use it to hop between them.
Pick one direction as positive and stick to it — if up is positive, g is negative all the way through.
Remember vvertical = 0 at the top — that single fact unlocks the maximum height.
⚠ Common mistakes
Swapping sin and cos when resolving — sketch the triangle to check
Using g in the horizontal direction — there’s no horizontal acceleration
Forgetting the horizontal velocity stays constant the whole flight
Not converting units, or losing the sign of g when up is taken as positive
Up next: Resistance in Fluids — what happens once we stop ignoring air resistance. Drag forces grow with speed, reshaping the neat parabola and leading towards ideas like terminal velocity.
Want this to actually click before the exam?
Book a free meeting and let’s work through the tricky bits together.
