Projectile Maximum Height

H = \frac{v_0^{2} \sin^{2}\theta}{2g}

Worked example: 30 m/s straight up → 45.887 m — press Try an example to run it live, then adjust anything.

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Constant used — built into this formula, no need to enter

g_{n} = 9.80665\ \text{m/s}^{2}

Standard acceleration of gravity · exact

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The height of a throw →

Grade 11Grade 11 Math — Functions & Applications

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Projectile Maximum Height explained

Only the vertical slice of the launch velocity, v₀ sin θ, fights gravity, and it is spent entirely by the time the projectile stops climbing — so the apex sits at v₀²sin²θ ⁄ 2g. Fire straight up at 30 m/s and you reach 30² ⁄ (2 × 9.80665) ≈ 45.9 m, about a 15-storey building. Galileo established the underlying parabola in Two New Sciences (1638) by rolling inked bronze balls off a table and marking where they struck — the first clean demonstration that horizontal and vertical motion proceed independently.

The common trap is squaring the sine rather than the whole term, or confusing this with the range formula's sin 2θ. Note how much steeper the sensitivity to angle is here than for range: a 60° launch reaches three times the height of a 30° launch at the same speed, yet lands at exactly the same spot. Basketball players exploit that trade — a high arc reaches the rim moving slowly and nearly straight down, which makes the hoop look bigger.

Projectile Maximum Height formula

H = \frac{v_0^{2} \sin^{2}\theta}{2g}

Where

$H$ = Maximum height (m)
$v_0$ = Launch speed (m/s)
$\theta$ = Launch angle (°)

Missing one of these? Work it out first, then come back

Maximum height — Gravitational Potential Energy (U = mgh), Area of a Triangle
Launch speed — Projectile Range on Level Ground, Projectile Time of Flight
Launch angle — Projectile Range on Level Ground, Projectile Time of Flight

Part of this formula set

The projectile motion equations

6 formulasRange, maximum height, time of flight and the two velocity components — the standard results for a projectile launched into still air.