Physics I: Mechanics › Kinematics › full formula sheet

R = v₀² sin(2θ) / gSay it: range equals launch speed squared, times sine of twice the angle, over g

Projectile range

How far a projectile flies on level ground — and the two-line derivation that reveals the magic 45°.

R is the horizontal range (m), v₀ launch speed, θ launch angle above horizontal, g = 9.8 m/s². Valid only for launch and landing at the same height, no air resistance.

Before this lesson: Projectile motion

Where it comes from

Shot-putters, golfers, and artillery crews all ask one question: what launch angle flies farthest? Beginners guess “as flat as possible” (more speed forward!) or “as high as possible” (more time aloft!). Both instincts are half-right — and the formula reconciles them exactly.

Before reading on: same launch speed — which flies farther, 30° or 60°? And what angle beats both? Lock in your guesses.
too flat (θ → 0°)
huge v₀x, but v₀y → 0 — lands almost immediately
All speed, no airtime.
too steep (θ → 90°)
huge airtime, but v₀x → 0 — lands at your feet
All airtime, no forward speed.
the sweet spot
θ = 45° — balances the two
And 30° vs 60°? Tie — complementary angles give identical ranges.

The formula below doesn’t just confirm this — it explains it, in one sine function.

Derivation

Range = (horizontal speed) × (flight time). We have both pieces from the projectile-motion lesson; multiply and simplify:

tflight
=
2v₀sinθ / g
Step 1 — flight time. On level ground the flight is symmetric: up-time = v₀y/g, doubled. (Landing back at launch height.)
R
=
v₀x · tflight = (v₀cosθ)(2v₀sinθ/g)
Step 2 — multiply. Horizontal speed is constant, so range is just v₀x times the whole flight.
=
v₀² · (2sinθcosθ) / g
Step 3 — collect. v₀ · v₀ = v₀²; the 2 joins the trig.
=
v₀² sin(2θ) / g
Step 4 — the double-angle identity. 2sinθcosθ = sin(2θ). The “2” inside the sine is not a fudge factor — it fell out of the algebra. ∎

Why 45°? sin(2θ) maxes at 1 when 2θ = 90°, i.e. θ = 45°. Why complementary angles tie? sin(2θ) = sin(180° − 2θ) = sin[2(90° − θ)] — so θ and 90° − θ (e.g. 30° and 60°) give the same range: one flies low and fast, the other high and slow, landing together.

How to use it

The procedure, every time:

  1. Confirm level ground. Launch height = landing height. A cliff or a raised tee breaks this formula — go back to the component equations.
  2. Compute sin(2θ) — double the angle first, then take the sine. Degrees mode on your calculator!
  3. Square v₀, multiply, divide by g. R = v₀² · sin(2θ) / 9.8.
  4. Sense-check. 45° should be your maximum; complementary pairs should tie.

Rearranged: how fast must I throw?

v₀ = √[Rg / sin(2θ)]solve for launch speed — “how hard to reach that far?”Say it: launch speed equals the square root of range times g over sine-two-theta

Want R = 100 m at 45°? v₀² = 100 · 9.8 / 1 = 980 → v₀ ≈ 31.3 m/s.

Common mistake: R = v₀² sinθ / g — forgetting to double the angle. At 30° that gives sin30° = 0.5 instead of sin60° = 0.866 — a 40%+ error from one dropped “2”.
Common mistake: using it for cliff launches. Fired from a 50 m cliff, the flight lasts longer than the symmetric t = 2v₀y/g — the formula undercounts the range. Level ground only.

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: v₀ = 30 m/s at 45°

  1. List givens. v₀ = 30 m/s, θ = 45°, level ground ✓
  2. Angle factor: sin(2·45°) = sin90° = 1.
  3. Compute. R = 30² · 1 / 9.8 = 900/9.8 ≈ 91.8 m.
  4. Sense-check. 45° is the max — any other angle at this speed flies shorter ✓.
Common mistake: calculator in radian mode: sin(90) with 90 radians ≈ 0.894, giving R ≈ 82 m. Degree mode for degree angles — check the little “DEG” indicator.
Your turn — v₀ = 20 m/s at 45°. Range?

Answer: 40.8 m. R = 400 · 1/9.8 ≈ 40.82 m.

Before reading on: v₀ = 25 m/s. Compute the range at 30° — then predict the range at 60° without computing. What does the formula say about the pair?

Example 2 — complementary angles: v₀ = 25 m/s at 30° (and 60°)

  1. At 30°: sin(60°) = 0.8660. R = 625 · 0.8660 / 9.8 = 541.27/9.8 ≈ 55.2 m.
  2. At 60°: sin(120°) = sin(60°) = 0.8660 — identical: 55.2 m.
  3. Why? sin(2θ) = sin(180° − 2θ). The 30° shot flies low and fast; the 60° shot lofts high and slow; they land together.
  4. And 45° beats both: 625/9.8 = 63.8 m > 55.2 m ✓.
Common mistake: “60° goes higher so it goes farther.” Higher, yes; farther, no — the lost horizontal speed exactly cancels the gained airtime. Symmetry, not intuition, decides.
Your turn — v₀ = 14 m/s at 60°. Range? (Then state the 30° range without computing.)

Answer: 17.3 m — and the 30° range is also 17.3 m. R = 196 · sin120°/9.8 = 196 · 0.8660/9.8 ≈ 17.32 m; complementary → tie.

Example 3 — solve for speed: R = 100 m at 45°

  1. Rearrange. v₀² = Rg/sin(2θ) = 100 · 9.8/1 = 980.
  2. Root. v₀ = √980 ≈ 31.3 m/s (about 113 km/h).
  3. Check. 31.3² · 1/9.8 = 979.7/9.8 = 100.0 m ✓.
  4. Reality note. A pro javelin throw (~90 m) needs roughly this speed — the numbers connect to real sport.
Common mistake: v₀ = Rg/sin(2θ) = 980 m/s — forgetting the square root. Units scream: R·g is m²/s², a squared speed. Root it.
Your turn — need R = 50 m at 45°. Launch speed?

Answer: 22.1 m/s. v₀² = 50 · 9.8 = 490; v₀ = √490 ≈ 22.14 m/s.

Example 4 — realistic angle: javelin, v₀ = 28 m/s at 42°

  1. Angle factor: sin(84°) ≈ 0.9945 — near the 45° max, as expected.
  2. Compute. R = 28² · 0.9945/9.8 = 784 · 0.9945/9.8 = 779.7/9.8 ≈ 79.6 m.
  3. Compare to 45°: 784/9.8 = 80.0 m — the 42° throw loses only 0.4 m. Near 45°, the sine curve is flat: small angle errors cost almost nothing.
  4. Real-world footnote. Actual javelin records use ~36° because aerodynamics (lift on the javelin) shifts the optimum below 45° — the formula is the no-air ideal.
Common mistake: sin(2·42°) computed as sin(42°) = 0.669 — nearly a third of the range vanishes. Double the angle before the sine, every time.
Your turn — v₀ = 22 m/s at 38°. Range?

Answer: 47.9 m. sin76° ≈ 0.9703; R = 484 · 0.9703/9.8 = 469.6/9.8 ≈ 47.92 m.

Memorization tips

  • Say it aloud: “range is v-naught squared, sine of two-theta, over g.” Three ingredients, one breath.
  • The 2θ is earned, not decorated: it came from 2sinθcosθ in the derivation. Doubling the angle before the sine is mandatory.
  • 45° is the king of level ground — sin(2θ) = 1 there. Complementary pairs (30°/60°) tie for second.
  • v₀² rules: double the speed → quadruple the range. Speed is the most powerful lever you have.
  • Level ground only. Cliff, hill, raised tee — the symmetry breaks and the formula undercounts. Fall back to components.
  • DEG mode. The #1 calculator error on this page: radians where degrees belong. Glance at the indicator before every sine.

Final challenge

Five mixed questions — basics, applications, and the traps, all in one. Score 5/5 and Projectile range 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 is 45 degrees the best angle for range?

Range is proportional to sin(2 theta), which peaks at 1 when 2 theta = 90 degrees, i.e. theta = 45 degrees. Physically it balances horizontal speed against airtime: flatter wastes airtime, steeper wastes forward speed.

Do 30 degrees and 60 degrees really give the same range?

Yes, on level ground with no air: sin(2 theta) = sin(180 - 2 theta), so theta and 90 - theta tie. The 30-degree shot flies low and fast; the 60-degree shot lofts high and slow; they land together.

When does the range formula NOT apply?

When launch and landing heights differ (cliffs, hills, raised tees) or air resistance matters. The derivation assumed symmetric flight t = 2 v0y/g, which breaks the moment the landing height changes.

Why does doubling the speed quadruple the range?

Because R is proportional to v0^2: faster launch means both more horizontal speed AND more airtime (higher arc), and the two multiply. (2v0)^2 = 4 v0^2.

Real javelin throwers use ~36 degrees, not 45. Why?

Aerodynamics. A real javelin gets lift, which shifts the optimum below 45 degrees. The formula is the no-air ideal — still the right starting point, and exact for dense balls at modest speeds.

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