Projectile Motion: Splitting It Into Horizontal and Vertical (AP & A-Level)

Quick answer: Projectile motion splits into two separate problems: horizontal motion at constant velocity (acceleration = 0) and vertical motion with constant downward acceleration g = 9.8 m/s^2. The only thing linking them is time — the same t applies to both directions. So resolve the launch velocity into a horizontal and a vertical component, run the suvat equations separately in each, and never let g touch the horizontal.

The first time projectile motion actually clicked for me was a demo my physics teacher did: she fired a ball horizontally off the bench and dropped a second ball from the same height at the exact same instant. They hit the floor together. That is the whole topic in one demo — moving sideways does not change how fast you fall.

Once I started treating a projectile as two boring straight-line motions stacked together instead of one curvy path, my marks jumped. It is the method AP and Cambridge A-Level examiners want: split, then solve.

Why horizontal stays constant but vertical speeds up

Gravity only pulls one way: down. The weight of the object, mg, acts vertically, so the only acceleration is vertical.

• Horizontally there is no force (once we ignore air resistance), so by Newton's first law there is no horizontal acceleration. The horizontal velocity you launch with is the horizontal velocity you keep — the whole way.
• Vertically, gravity gives a constant acceleration of g = 9.8 m/s^2 downwards, whatever the object is doing sideways.

The two directions are independent — the only thing they share is the clock. The same t appears in both, and that shared time is how you cross from one direction to the other.

Your toolkit: suvat in each direction

The suvat equations (u = initial velocity, v = final velocity, a = acceleration, s = displacement, t = time):

• v = u + at
• s = ut + (1/2)at^2
• v^2 = u^2 + 2as
• s = (1/2)(u + v)t

Set them up in two columns first. Horizontal: a = 0, so it collapses to just s = ut. Vertical: a = g, so you use the full set. Decide which direction is positive and keep it for the whole question.

Worked example 1 — horizontal launch

A ball leaves the edge of a table 1.25 m high, moving horizontally at 3.0 m/s. How long is it in the air, and how far from the table does it land? Take g = 9.8 m/s^2 and ignore air resistance.

Step 1 — split the motion. Horizontal: u = 3.0 m/s, a = 0. Vertical: u = 0 (launched horizontally, so no initial vertical speed), a = 9.8 m/s^2 downward, s = 1.25 m.

Step 2 — use the vertical direction to find the time. The fall controls the time, not the sideways speed.

s = ut + (1/2)at^2, so 1.25 = 0 + (1/2)(9.8)t^2, giving 1.25 = 4.9t^2, then t^2 = 0.255 and t = 0.51 s.

Step 3 — feed that time into the horizontal direction for the range. s = ut = 3.0 × 0.51 = 1.5 m.

So the ball is airborne about 0.51 s and lands roughly 1.5 m away. Notice the 3.0 m/s never entered the time calculation — a ball simply dropped from 1.25 m would land in the same 0.51 s. That is the two-motions idea in action.

Worked example 2 — launch at an angle

A projectile is launched at 20 m/s at 30 degrees above the horizontal on level ground. Find the time of flight, the maximum height, and the range. Take g = 9.8 m/s^2.

Step 1 — resolve into components. u across = 20 cos30 = 17.3 m/s. u up = 20 sin30 = 10.0 m/s.

Step 2 — time of flight from the vertical. At the top the vertical velocity is zero: 0 = 10.0 - 9.8t, so time to the top is 1.02 s. On level ground the way down mirrors the way up, so total time of flight is 2 × 1.02 = 2.04 s.

Step 3 — maximum height. v^2 = u^2 - 2gH, so 0 = 10.0^2 - 2(9.8)H, giving H = 100 / 19.6 = 5.1 m.

Step 4 — range, horizontal distance over the full flight. s = ut = 17.3 × 2.04 = 35.3 m.

Time of flight and range: the shortcuts, and when they break

For a projectile that lands at the same height it launched from, you can quote:

• Time of flight: T = 2u sinθ / g
• Maximum height: H = (u sinθ)^2 / (2g)
• Range: R = u^2 sin(2θ) / g, largest at θ = 45 degrees

One warning that catches people every year: these only work when the landing height equals the launch height. Fire something off a cliff so it lands lower and R = u^2 sin(2θ) / g is simply wrong — split the motion again and use the vertical suvat with the real displacement (a negative s if it ends below the start). If the algebra gets messy, drop the line into Math solver to check each rearrangement.

Common mistakes that cost marks

1. Using the whole launch speed instead of its component. The 20 m/s is not the vertical u — resolve first. Putting the resultant straight into a vertical equation is the single most common error.
2. Letting g into the horizontal. Horizontal acceleration is zero. No g sideways, ever.
3. Sign chaos. Choose up = positive (or down = positive) before you start and keep it. If up is positive, then g = -9.8 and a downward displacement is negative.
4. Thinking a horizontal launch has vertical u. For a ball rolled off a table, the initial vertical velocity is 0.
5. Using the range formula at a different landing height. Only valid for equal launch and landing heights.
6. Forgetting to state the model. Both AP and Cambridge assume you ignore air resistance and treat the object as a particle — a mark can hang on saying so.

Test yourself

1. (AP) A stone is thrown horizontally at 6.0 m/s from a cliff 20 m high. How long until it lands, and how far from the base? (g = 9.8)
2. (A-Level) A ball is kicked at 15 m/s at 40 degrees to the horizontal on level ground. Find the maximum height. (g = 9.8)
3. (Concept) Two balls leave the same height at the same instant — one dropped, one thrown horizontally at 10 m/s. Which lands first?

Quick answers: (1) Vertical: 20 = 4.9t^2, so t = 2.02 s; horizontal: 6.0 × 2.02 = 12.1 m. (2) u up = 15 sin40 = 9.64 m/s, so H = 9.64^2 / 19.6 = 4.7 m. (3) They land together — the vertical motion is identical, and horizontal speed does not change the fall time.

Want your working checked in your board's style? Paste it into Math solver and it will flag exactly where a sign or component slipped. To build the intuition first, ask Explain a concept to walk through the two-directions idea, or drill the split on Quiz me until it is automatic.

FAQ

Why does a projectile follow a parabola?

Because horizontal position grows steadily with time (distance proportional to t) while vertical position changes with t^2 (the (1/2)at^2 term). Combine a straight-line term with a squared term and the shape you trace is a parabola.

Does the mass of the projectile affect its path?

No, as long as you ignore air resistance. g is the same for every mass, so a heavy and a light object launched identically follow the same path and land together. Mass only matters once air resistance is included.

What launch angle gives the maximum range?

On level ground, 45 degrees. There is a neat symmetry too: angles the same amount above and below 45, such as 30 and 60 degrees, give the same range.

Should I use g = 9.8 or 9.81?

Check your paper. AP typically uses 9.8 m/s^2, while Cambridge A-Level (9702) usually prints g = 9.81 m/s^2 on the data sheet. Use whatever the exam states and stay consistent through the whole question.

In short: treat a projectile as two independent motions sharing one clock — constant velocity across, constant acceleration g down. Resolve the launch velocity into components, use the vertical direction for the time, then use that time in the horizontal for the range. Keep g out of the horizontal, keep signs consistent, and only use the range formula when it lands at the height it left. Do that and projectiles become some of the most predictable marks on the paper.