Sports & Motion Analytics
Intermediate

Projectile Max Height

Peak height reached by a launched projectile.

Formula

H=v2sin⁡2θ2gH = \frac{v^2 \sin^2\theta}{2g}

Variables

HMax height (m)
vLaunch speed (m/s)
θLaunch angle
gGravity 9.81 m/s²

Example

20 m/s at 60°: H = 400×0.75÷19.62 ≈ 15.3 m

How it works

This formula gives the highest point a projectile reaches when it is launched and then moves under gravity alone, assuming air resistance is negligible. It comes from splitting the launch velocity into vertical and horizontal parts: only the vertical component, v sinθ, determines how high the object rises, and gravity slows that upward motion until the vertical speed becomes zero at the top. Using the constant-acceleration equation for vertical motion leads to H = (v^2 sin^2θ) / (2g).

When to use it

Use this when analyzing the peak height of a kicked ball, thrown object, launched projectile, or any motion where the object follows ideal projectile motion. It is especially useful in sports science, physics problems, and trajectory calculations where launch speed and angle are known.

Worked example, step by step

A javelin is launched at 18 m/s and 45°. Step 1: Square the speed: v^2 = 18^2 = 324 Step 2: Find sinθ and square it: sin45° = 0.7071, so sin^2 45° ≈ 0.5 Step 3: Multiply the top: 324 × 0.5 = 162 Step 4: Multiply the bottom: 2g = 2 × 9.81 = 19.62 Step 5: Divide: H = 162 ÷ 19.62 ≈ 8.25 m Final answer: The maximum height is about 8.3 m.

Common mistakes to avoid

  • Using cosθ instead of sinθ, which gives the horizontal component instead of the vertical one
  • Forgetting to square the sine term or the speed, which changes the result a lot
  • Using a different value of g without checking units, especially mixing m/s² with other gravity values

Did You Know?

A football kicked at 60° climbs higher but lands shorter than one at 30° with the same speed.

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Reviewed & maintained by the FormulaWon editorial teamLast reviewed September 2026