Physics I: Mechanics › Kinematics › full formula sheet

x = (v₀cosθ)t     y = (v₀sinθ)t − ½gt²Say it: horizontal position grows steadily with the horizontal launch speed; vertical position follows free fall with the vertical launch speed

Projectile motion

Two motions for the price of one — horizontal cruising and vertical free fall, completely independent, sharing only the clock.

θ is the launch angle above the horizontal. v₀cosθ is the horizontal launch velocity (constant), v₀sinθ the vertical launch velocity. Horizontal: ax = 0. Vertical: ay = −g. Air resistance neglected.

Before this lesson: Free fall

Where it comes from

A cannonball arcs; a long jumper sails; a basketball swishes. Every one of these looks complicated — a curve! — until Galileo’s key insight splits it in two: horizontal motion and vertical motion don’t talk to each other. Sideways, the projectile cruises at constant velocity (nothing pushes it). Vertically, it free-falls. The curve is just the two motions drawn together.

Before reading on: a rifle is fired horizontally from shoulder height; at the same instant, a bullet is simply dropped from the same height. Which hits the ground first? Commit to an answer before reading on.
fired bullet
y = −½gt²   (v₀y = 0)
Its horizontal speed is enormous — and irrelevant to the fall.
dropped bullet
y = −½gt²   (v₀y = 0)
Identical vertical equation. They land simultaneously — the most famous demo in mechanics.

That demo is the whole theory in miniature: the vertical motion never knew the horizontal motion existed. The equations below just write that independence down.

Derivation

Apply Newton’s second law per axis. Horizontally there are no forces (ignoring air); vertically there is gravity. Then integrate, axis by axis:

ax = 0,   ay = −g
Step 1 — the physics, per axis. No horizontal force → no horizontal acceleration. Gravity pulls down only.
vx = v₀x,   vy = v₀y − gt
Step 2 — integrate once. v = v₀ + at, per axis. Horizontal velocity is frozen at its launch value forever.
v₀x = v₀cosθ,   v₀y = v₀sinθ
Step 3 — decompose the launch. Adjacent over hypotenuse: horizontal gets cosine, vertical gets sine. (θ = 0 → all horizontal; θ = 90° → all vertical. Check the extremes!)
x = (v₀cosθ)t
,
y = (v₀sinθ)t − ½gt²
Step 4 — integrate again. The hero equations. Two independent motions, one shared clock t. ∎

Eliminate t for the trajectory shape: t = x/(v₀cosθ) gives y = x·tanθ − [g/(2v₀²cos²θ)]x² — a parabola. That’s why every projectile path (no air) is a parabolic arc.

How to use it

The procedure, every time:

  1. Decompose first. v₀x = v₀cosθ, v₀y = v₀sinθ. Write both down before touching any equation.
  2. Split the question by axis. “How far?” is horizontal (needs the flight time). “How high / how long in the air?” is vertical (free fall with v₀y).
  3. Let the vertical set the clock. Flight time almost always comes from the y-equation (lands when y returns to launch height, or hits a given height).
  4. Carry t to the horizontal. x = v₀x · t finishes the job.

The two special launches

θ = 0° (horizontal launch):   x = v₀t,   y = −½gt²fired level — the dropped-bullet demoSay it: sideways it cruises, downward it free-falls from rest
θ = 90° (straight up):   x = 0,   y = v₀t − ½gt²pure free fall — the previous lesson
Common mistake: putting −½gt² in the x-equation too (“gravity affects everything”). Gravity is vertical only. The x-equation has no g, no t² — ever.
Common mistake: swapping sine and cosine. Check the extremes: θ = 0 should give all-horizontal (cos0 = 1 ✓), θ = 90° all-vertical (sin90 = 1 ✓). If your assignment fails the extremes, swap them.

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the basic move: soccer ball, v₀ = 20 m/s at 30°, at t = 1 s

  1. Decompose. v₀x = 20·cos30° = 20·0.8660 = 17.32 m/s; v₀y = 20·sin30° = 20·0.5 = 10 m/s.
  2. Horizontal: x = 17.32 · 1 = 17.32 m.
  3. Vertical: y = 10·1 − 4.9·1² = 10 − 4.9 = 5.1 m.
  4. Picture it. After 1 s the ball is 17.3 m downfield and 5.1 m up — climbing (vy = 10 − 9.8 = +0.2 m/s, barely still rising).
Common mistake: x = 20·1 = 20 m — using the full v₀ horizontally. Only the horizontal component drives x; the full speed overcounts it.
Your turn — v₀ = 15 m/s at 45°; position at t = 1 s?

Answer: x = 10.61 m, y = 5.71 m. v₀x = v₀y = 15 · 0.7071 = 10.61 m/s; x = 10.61 m; y = 10.61 − 4.9 = 5.71 m.

Example 2 — time to the peak: same ball (v₀y = 10 m/s)

  1. Peak means vy = 0. 0 = v₀y − gt = 10 − 9.8t.
  2. Solve. t = 10/9.8 ≈ 1.02 s.
  3. Height there: y = 10·1.0204 − 4.9·(1.0204)² = 10.204 − 5.102 = 5.10 m.
  4. Note. The horizontal motion is irrelevant to the peak — this is pure free fall with v₀ = 10 m/s upward.
Common mistake: t = v₀/g = 20/9.8 = 2.04 s — using the full launch speed instead of the vertical component. Only v₀y fights gravity.
Your turn — v₀ = 14 m/s at 30°. Time to peak? Peak height?

Answer: 0.714 s; 2.5 m. v₀y = 7 m/s; t = 7/9.8 ≈ 0.714 s; y = 7 · 0.714 − 4.9 · 0.510 = 2.5 m.

Example 3 — horizontal launch: cliff 50 m, v₀x = 12 m/s

  1. Decompose. θ = 0°: v₀x = 12 m/s, v₀y = 0. The vertical is a pure drop.
  2. Vertical sets the clock. Lands when y = −50: −50 = −4.9t² → t² = 50/4.9 = 10.204 → t ≈ 3.19 s.
  3. Carry to horizontal. x = 12 · 3.1944 ≈ 38.3 m from the cliff base.
  4. The demo, quantified. A dropped rock from the same cliff also takes 3.19 s — the 12 m/s sideways changed nothing vertically.
Common mistake: t = 50/12 = 4.17 s — dividing the height by the horizontal speed. Height is vertical business; horizontal speed never enters the fall time.
Your turn — cliff 45 m, fired level at 10 m/s. Time to land? Horizontal distance?

Answer: 3.03 s; 30.3 m. t² = 45/4.9 = 9.184 → t ≈ 3.03 s; x = 10 · 3.03 = 30.3 m.

Before reading on: v₀ = 25 m/s at 37°, landing back at launch height. The vertical launch speed is about 15 m/s — so roughly how long is the flight? And will the range exceed 60 m? Estimate first.

Example 4 — full flight on level ground: v₀ = 25 m/s at 37°

  1. Decompose. v₀y = 25·sin37° = 25·0.6018 = 15.05 m/s; v₀x = 25·cos37° = 25·0.7986 = 19.97 m/s.
  2. Flight time (vertical, symmetric). t = 2v₀y/g = 2·15.05/9.8 ≈ 3.07 s.
  3. Range (horizontal). R = 19.97 · 3.07 ≈ 61.3 m.
  4. Cross-check with the range formula (next lesson): R = v₀²sin(2θ)/g = 625·sin74°/9.8 = 625·0.9613/9.8 = 61.3 m ✓
Common mistake: R = v₀ · t = 25 · 3.07 = 76.8 m — using full speed for the horizontal leg. The range formula exists precisely to stop this; only v₀x crosses the field.
Your turn — v₀ = 20 m/s at 53°, level ground. Flight time? Range?

Answer: 3.26 s; 39.2 m. v₀y = 20 · 0.7986 = 15.97 m/s; t = 2 · 15.97/9.8 ≈ 3.26 s; v₀x = 20 · 0.6018 = 12.04 m/s; R = 12.04 · 3.26 ≈ 39.2 m.

Memorization tips

  • Say it aloud: “sideways it cruises, up-down it free-falls — one clock for both.” Independence is the whole lesson.
  • Decompose before you compute. v₀x and v₀y on paper first; every later step reads off those two numbers.
  • Extremes check: θ = 0 → all horizontal; θ = 90° → all vertical. If your sine/cosine assignment fails an extreme, swap them.
  • Vertical sets the clock. Flight time, time to peak, hang time — all vertical questions. Solve y first, carry t to x.
  • The path is a parabola (no air): y = x·tanθ − [g/(2v₀²cos²θ)]x². The minus-x² term is gravity bending the straight line down.
  • No g in the x-equation. Ever. If you wrote one, you mixed the axes.

Final challenge

Five mixed questions — basics, applications, and the traps, all in one. Score 5/5 and Projectile motion is yours.

← Back to the Physics I formula sheet

How to learn a formula here

  1. Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
  2. Work the examples with the answers covered, then uncover one step at a time and compare.
  3. Finish with the final challenge — five mixed questions including the classic traps.
  4. Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

Why are horizontal and vertical motion independent?

Because forces don't mix axes: with no air resistance, there is no horizontal force (so a_x = 0 and v_x stays constant) while gravity acts purely vertically (a_y = -g). Each axis follows its own equation, sharing only the time t.

Which hits first: a fired bullet or a dropped bullet?

They land together (same height, no air). Both have v0y = 0, so both obey y = -1/2 g t^2 vertically. The fired bullet's huge horizontal speed never enters the vertical equation.

How do I split the launch velocity into components?

v0x = v0 cos(theta) (adjacent/horizontal) and v0y = v0 sin(theta) (opposite/vertical), with theta above the horizontal. Check extremes: theta = 0 gives all-horizontal, theta = 90 degrees all-vertical.

What is the trajectory shape?

A parabola: eliminating t gives y = x tan(theta) - [g/(2 v0^2 cos^2(theta))] x^2. The x^2 term with its negative coefficient is gravity bending the path downward.

How do I find the flight time?

Solve the vertical equation for when the projectile reaches the landing height. On level ground it's t = 2 v0 sin(theta)/g (twice the time to the peak, by symmetry). Then range = v0x times that t.

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