Kick a football into the air, fire a cannon, spray a garden hose, throw a set of keys across the room — every one of them traces the same graceful arc, rising, curving over, and falling back to earth. That arc is a projectile: any object moving under gravity alone once it has been launched, with no engine, no thrust, and (for now) no air resistance to worry about.

A projectile seems to do two things at once — it travels along the ground and it moves up and down — and that sounds complicated. The single beautiful idea of this whole page is that it is not complicated, because those two motions are completely independent. The sideways motion knows nothing about the up-and-down motion, and the up-and-down motion knows nothing about the sideways one. Split the problem in two, treat each direction with the suvat equations you already know, and a curving arc becomes two easy straight-line problems.

Split the launch velocity into two directions

A projectile is usually launched at some speed u and some angle \theta above the horizontal. Before doing anything else, resolve that single launch velocity into a horizontal part and a vertical part — exactly the vector-splitting from scalars and vectors. The launch arrow is the hypotenuse of a right-angled triangle:

From this instant on we treat u_x and u_y as two separate starting velocities for two separate problems. The horizontal one is laughably easy; the vertical one is just free fall. Let's take them one at a time.

Horizontal: nothing pushes it sideways

Once the object is in the air, what force acts on it horizontally? Ignoring air resistance — none. Gravity pulls straight down, so it has no sideways component at all. No horizontal force means no horizontal acceleration, and no horizontal acceleration means the sideways velocity never changes. The projectile drifts across at a constant u_x for the whole flight.

Constant velocity is the simplest motion there is — distance is just speed times time:

That is the entire horizontal story. No g, no acceleration, no squared term — the horizontal distance simply ticks up in equal steps, second after second.

Vertical: it's just free fall

Vertically, the projectile feels the one force that never lets go: gravity. It has a constant downward acceleration of magnitude g \approx 9.8\ \text{m/s}^2, so the vertical motion is a standard constant-acceleration problem — the very thing the suvat equations were made for. Taking up as positive, the initial vertical velocity is u_y = u\sin\theta and the acceleration is a = -g (negative because gravity points down):

These are exactly the suvat equations, with u_y in place of u and -g in place of a. As the object rises, gravity steadily eats away its upward speed; at the top the upward speed has been reduced to zero; then gravity keeps pulling and it speeds up on the way down. Same equation throughout — up, over, and down.

Time is the secret link

If the two motions are truly independent, what ties them together into one arc? The answer is the one quantity they share: time t. The clock runs at the same rate for the sideways motion and the up-and-down motion — every second that passes vertically is the same second that passes horizontally.

So the recipe for any projectile problem is always the same:

• Use the vertical equations (with g) to find a time — the time to reach the top, or the time to hit the ground.
• Feed that same time into the horizontal equation x = u_x t to find how far it has travelled sideways.

Vertical motion tells you when; horizontal motion tells you how far. Get the time from one direction, spend it in the other. That's the whole trick.

Worked example: fired horizontally off a cliff

A ball is thrown horizontally at u = 15\ \text{m/s} from the top of a cliff h = 20\ \text{m} high. Take g = 9.8\ \text{m/s}^2. How long is it in the air, and how far from the base of the cliff does it land?

"Horizontally" means the launch angle is 0^\circ, so all the launch speed is horizontal and there is no initial vertical velocity: u_x = 15\ \text{m/s} and u_y = 0.

Step 1 — vertical, to get the time. It starts with no downward speed and falls 20\ \text{m}. Using h = \tfrac{1}{2}g\,t^2:

Step 2 — horizontal, using that same time. The sideways speed is a constant 15\ \text{m/s}, so

It lands about 30\ \text{m} from the foot of the cliff. Notice how the vertical calculation never once mentioned the 15\ \text{m/s}, and the horizontal calculation never once mentioned gravity — the two directions really were handled completely separately, meeting only through the shared 2.0\ \text{s}.

Worked example: launched at an angle

Now a projectile is launched from level ground at u = 20\ \text{m/s}, at \theta = 30^\circ above the horizontal, with g = 9.8\ \text{m/s}^2. Find its time of flight, its maximum height, and its range.

Step 1 — resolve the launch velocity.

Step 2 — time to the top. At the highest point the vertical velocity is momentarily zero. Using v_y = u_y - g\,t with v_y = 0:

Step 3 — total time of flight. By symmetry, the trip down takes exactly as long as the trip up, so the total time in the air is double: T = 2t_{\text{up}} \approx 2.04\ \text{s}.

Step 4 — maximum height. Using H = \dfrac{u_y^{\,2}}{2g}:

Step 5 — range. Spend the whole flight time in the horizontal equation:

Why the dropped ball and the fired ball land together

Here is the classic demonstration that makes the independence idea unforgettable. Hold two identical balls at the same height. Drop one straight down. At the very same instant, fire the other one horizontally, as fast as you like. Which hits the floor first?

They land at exactly the same moment. The fired ball goes shooting off across the room, yet it hits the ground at the same instant as the one that simply fell. Why? Because their vertical motions are identical: both start with zero vertical velocity and both fall under the same g. The fired ball's extra horizontal speed does nothing whatsoever to how fast it falls — the horizontal and vertical motions are independent. The bullet fired dead level from a gun and the bullet dropped from your hand at the same height reach the ground together (on flat ground, ignoring air resistance and the Earth's curvature).

Play with a trajectory

Below is a projectile fired from the origin over level ground (with g = 9.8\ \text{m/s}^2). Drag the launch speed and launch angle and watch the arc respond. The dashed lines mark the maximum height reached at the top of the arc, and the dot on the ground marks the range where it lands. The live readout shows the resolved calculations.

• Fix the speed and sweep the angle: the range grows to a maximum at 45^\circ, then shrinks again — and angles either side of 45^\circ (say 30^\circ and 60^\circ) give the same range.
• Crank the angle high (near 75^\circ): a tall, short lob. Flatten it low (near 15^\circ): a fast, shallow skim.
• Every arc is a parabola — the signature curve of "constant sideways speed, constant downward pull."

The path is a parabola — and air resistance bends it

Why a parabola exactly? Combine the two motions by eliminating time. Horizontally x = u_x t, so t = x / u_x. Substitute that into the vertical equation y = u_y t - \tfrac{1}{2}g t^2:

That is y = (\text{constant})x - (\text{constant})x^2 — a quadratic in x, and the graph of a quadratic is a parabola. The symmetry of the parabola is why the way up mirrors the way down.

In the real world, air resistance spoils this tidy picture. Because drag always pushes against the motion, it saps both the horizontal and the vertical speed, so a real thrown ball travels a shorter range, doesn't climb as high, and — crucially — its path becomes asymmetric: the descent is steeper and slower than the ascent, no longer a perfect mirror image. It is closely related to air resistance. The clean parabola is the idealised, no-air answer — the right place to start, and a very good approximation for heavy, slow, streamlined objects.