Projectile Maximum Height

H=v02sin2θ2gH = \frac{v_0^{2} \sin^{2}\theta}{2g}

Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!

Learning zone

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
H=v02sin2θ2gH = \frac{v_0^{2} \sin^{2}\theta}{2g}
Where
  • HH= Maximum height
  • v0v_0= Launch speed
  • θ\theta= Launch angle