Physics I: Mechanics › Kinematics › full formula sheet

H = (v₀sinθ)² / (2g)Say it: max height equals the vertical launch speed squared, over twice g

Projectile max height

How high a projectile climbs — and why the horizontal half of the launch is completely irrelevant to the answer.

H is the peak height above the launch point, v₀sinθ the vertical launch speed, g = 9.8 m/s². Launch and peak measured from the same launch height; no air resistance.

Before this lesson: Projectile motion

Where it comes from

A basketball arcs toward the hoop; a firework shell climbs before it bursts; a high jumper’s center of mass peaks and falls. In each case someone wants to know how high — and the answer ignores half the motion. The horizontal speed could be 5 m/s or 500 m/s; the peak height doesn’t care.

Before reading on: two balls launched at the same speed — one at 30°, one at 60°. Which peaks higher? And what happens to the peak if you double the launch speed?
30°
v₀y = v₀ · 0.5
Half the speed fights gravity.
60°
v₀y = v₀ · 0.866
More vertical speed → higher peak. And doubling v₀? The formula squares it — 4× the height.

The peak is a purely vertical event: the instant the vertical velocity hits zero. Everything below derives from that one fact.

Derivation

At the peak, the vertical velocity is momentarily zero. Feed that into the time-free equation (which never asks for t):

vy²
=
v₀y² − 2gH
Step 1 — time-free, vertical. v² = v₀² + 2aΔy with a → −g and Δy → H. No time anywhere — exactly what a “how high” question wants.
0
=
v₀y² − 2gH
Step 2 — the peak condition. At the top the projectile stops rising: vy = 0. (Horizontal velocity is still v₀x — the projectile is not stationary!)
H
=
v₀y² / (2g) = (v₀sinθ)² / (2g)
Step 3 — solve and substitute. 2gH = v₀y², and v₀y = v₀sinθ. Done. ∎

Energy shortcut: ½mv₀y² = mgH (vertical kinetic energy converts fully to potential at the peak) gives the same H — the mass cancels. Two roads, one formula.

How to use it

The procedure, every time:

  1. Extract the vertical launch speed: v₀y = v₀sinθ. This is the only part of the launch that matters.
  2. Square it, divide by 2g = 19.6. H = v₀y²/19.6.
  3. Remember what H measures: height above the launch point. Add the launch height for height above ground.
  4. Sanity-check the scaling: double v₀y → 4× H. If your numbers don’t respect that, recheck.

Special cases

θ = 90° (straight up):   H = v₀²/(2g)sin90° = 1 — the free-fall peak formulaSay it: straight up, the whole launch speed fights gravity
θ = 0° (level):   H = 0no vertical launch speed, no climb — the dropped-bullet case
Common mistake: H = v₀²/(2g) — using the full launch speed. At 30° that overcounts by 4× (sin²30° = 0.25). Only the vertical component climbs.
Common mistake: “the projectile is stationary at the peak, so H comes from v = 0” — then using v₀x = 0 too. At the peak vy = 0 but vx = v₀x still; the projectile is moving horizontally at the top.

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: straight up at 30 m/s

  1. List givens. v₀ = 30 m/s, θ = 90° → v₀y = 30 m/s.
  2. Apply H = v₀y²/(2g). = 30²/19.6 = 900/19.6.
  3. Compute. ≈ 45.9 m — about 15 stories.
  4. Cross-check via time. tup = 30/9.8 = 3.06 s; y = 30·3.06 − 4.9·9.37 = 91.8 − 45.9 = 45.9 m ✓
Common mistake: H = 30/19.6 = 1.53 m — forgetting to square. Units catch it: v/g is seconds, not meters. Squared speed over g is meters.
Your turn — straight up at 20 m/s. Max height?

Answer: 20.4 m. H = 400/19.6 ≈ 20.41 m.

Example 2 — angled: v₀ = 25 m/s at 30°

  1. Vertical component. v₀y = 25 · sin30° = 25 · 0.5 = 12.5 m/s.
  2. Square and divide. H = 12.5²/19.6 = 156.25/19.6 ≈ 7.97 m.
  3. Compare with straight up at the same speed: 625/19.6 = 31.9 m — the 30° launch reaches exactly 1/4 of that (sin²30° = 0.25) ✓.
Common mistake: H = 25²/19.6 = 31.9 m — the full-speed error. Ask: “would a 30° lob really match a vertical rocket?” No — the sine-squared factor is the discount.
Your turn — v₀ = 18 m/s at 30°. Max height?

Answer: 4.13 m. v₀y = 9 m/s; H = 81/19.6 ≈ 4.13 m.

Before reading on: you need to clear a 3 m wall, launching at 60°. Roughly how fast must the ball leave — faster or slower than 10 m/s? Ballpark it, then solve.

Example 3 — solve for speed: clear H = 3 m at θ = 60°

  1. Rearrange. v₀²sin²θ = 2gH → v₀² = 2gH/sin²θ.
  2. Plug in. v₀² = 2·9.8·3/(sin60°)² = 58.8/0.75 = 78.4.
  3. Root. v₀ = √78.4 ≈ 8.85 m/s.
  4. Check. v₀y = 8.85·0.866 = 7.67 m/s; H = 58.8/19.6 = 3.0 m ✓.
Common mistake: v₀ = 2gH/sinθ — dropping both the square root and the square on the sine. Two missing operations compound: always rearrange the full H = (v₀sinθ)²/(2g) line by line.
Your turn — need H = 5 m at θ = 45°. Launch speed?

Answer: 14 m/s. v₀² = 2 · 9.8 · 5/0.5 = 196; v₀ = 14 m/s exactly.

Example 4 — the scaling law: v₀ 10→20 m/s at 45°

  1. At 10 m/s: v₀y = 10·sin45° = 7.071 m/s; H = 7.071²/19.6 = 50/19.6 ≈ 2.55 m.
  2. At 20 m/s: v₀y = 14.142 m/s; H = 200/19.6 ≈ 10.20 m.
  3. Ratio: 10.20/2.55 = 4.0× — doubling speed quadruples height.
  4. Why it matters. A 10% faster throw buys 21% more height (1.1² = 1.21) — small speed edges pay off quadratically.
Common mistake: “twice the speed, twice the height.” The square says otherwise — and the same scaling governs range, stopping distance, and crater size. Learn it once, use it everywhere.
Your turn — straight up: v₀ 15→30 m/s. Both heights? Ratio?

Answer: 11.48 m and 45.92 m — ratio 4. H = 225/19.6 ≈ 11.48 m; H = 900/19.6 ≈ 45.92 m; 45.92/11.48 = 4.0.

Memorization tips

  • Say it aloud: “max height is vertical launch speed squared over twice g.” Vertical. Squared. Twice-g.
  • Only v₀sinθ climbs. The horizontal component is a spectator at the peak — it never enters H.
  • 2g = 19.6 — memorize the denominator. H = (vertical speed)² / 19.6, done.
  • Peak ≠ stopped: vy = 0 but vx = v₀x. The projectile cruises horizontally across its own summit.
  • Square-law reflex: ×2 speed → ×4 height; ×3 → ×9. Apply it as a sanity check on every answer.
  • H is from launch height. Launched from a cliff or a hand? Add the launch height for ground clearance.

Final challenge

Five mixed questions — basics, applications, and the traps, all in one. Score 5/5 and Projectile max height 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 doesn't horizontal speed affect the max height?

Height is a vertical event: the peak is where vertical velocity hits zero. Horizontal motion has no vertical component and gravity acts only vertically, so v0x never enters the equation H = (v0 sin theta)^2/(2g).

Is the projectile stationary at the peak?

No — only its vertical velocity is zero there. The horizontal velocity v0x persists unchanged, so it cruises horizontally across the summit. 'Momentarily not rising' is not 'stopped'.

Why does doubling the speed quadruple the height?

Because H is proportional to v0y^2. (2v)^2 = 4v^2 — the same square law behind range quadrupling and stopping distances. It's one of the most-used scaling facts in mechanics.

How is this related to energy?

At the peak, the vertical kinetic energy 1/2 m v0y^2 has fully converted to gravitational potential mgH. Setting them equal gives H = v0y^2/(2g) — the mass cancels, so it's the same formula.

What if the projectile is launched from a height?

H measures height above the launch point. For height above the ground, add the launch height: H_ground = H + h_launch. (For the full trajectory from a height, solve the y-equation directly.)

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