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Physics · Mechanics

Projectile Motion Calculator

From launch speed and angle, get the range, maximum height, and flight time.

m/s
°
m/s²
Examples — tap to load
Range
40.775m

Max height: 10.194 m · Flight time: 2.883 s

Trajectory
Projectile trajectory arc10 m0 m0 m41 m

On level ground the range is R = v₀² sin(2θ) ÷ g. Launching at v₀ = 20 m/s and θ = 45° with g = 9.81 m/s² gives R = 20² × sin(90°) ÷ 9.81 = 400 ÷ 9.81 = 40.78 m, the farthest distance for that speed.

How projectile motion works

Once a projectile leaves your hand, the only force on it is gravity (ignoring air resistance). The trick is that horizontal and vertical motion are completely independent: the horizontal speed never changes, while the vertical speed slows, stops, and reverses. Splitting the launch velocity into those two components is the whole game, and the vertical part follows the same SUVAT equations of motion.

R = v₀² sin(2θ) ÷ g  ·  H = (v₀ sinθ)² ÷ 2g  ·  T = 2v₀ sinθ ÷ g

v₀ = launch speed, θ = launch angle, g = gravity (level ground)

Worked example

A ball launched at v₀ = 20 m/s, θ = 45°, on Earth (g = 9.81 m/s²):

  1. 1
    Find the flight time. T = 2 × 20 × sin45° ÷ 9.81 = 2 × 14.142 ÷ 9.81 = 2.883 s in the air.
  2. 2
    Find the maximum height. H = (20 × sin45°)² ÷ (2 × 9.81) = 14.142² ÷ 19.62 = 200 ÷ 19.62 = 10.194 m.
  3. 3
    Find the range. R = 20² × sin(90°) ÷ 9.81 = 400 × 1 ÷ 9.81 = 40.775 m horizontally.

Range vs launch angle (v₀ = 20 m/s, g = 9.81 m/s²)

Range depends on sin(2θ), so 30° and 60° give the same distance; 45° is the peak.

Angle θsin(2θ)Range R
15°0.50020.39 m
30°0.86635.31 m
45°1.00040.77 m
60°0.86635.31 m
75°0.50020.39 m

Assumptions and common mistakes

This is the drag-free model. Air resistance is ignored, which is standard for introductory physics but means real-world balls fall short of these numbers.

It assumes level ground. The range formula above assumes the projectile lands at the same height it was launched from. A cliff or raised platform needs the fuller equations.

The 45° rule: on level ground, 45° gives the maximum range for a fixed speed because sin(2θ) peaks at sin(90°) = 1. Note that maximum height, by contrast, keeps growing all the way to 90°.

Does it include air resistance?
No — this is the idealized, drag-free model, which is standard for introductory physics. Real projectiles experience drag and fall somewhat short of these values.
What angle gives the longest range?
45° on level ground, for a fixed launch speed. Because range scales with sin(2θ), pairs like 30° and 60° travel the same distance.
Can I change gravity?
Yes — adjust g to model other planets (Moon 1.62, Mars 3.72 m/s²) or keep 9.81 m/s² for Earth.
Why does the ball land at the same height it started?
The range formula assumes level ground. If you launch from a height, the flight time and range increase — that case needs the full kinematic equations.
How is maximum height related to angle?
Height depends on sinθ, not sin(2θ), so it keeps rising as the angle approaches 90° — a straight-up throw reaches the greatest height but zero range.
How do I calculate the range from speed and angle?
Use R = v₀² sin(2θ) ÷ g. At v₀ = 20 m/s, θ = 45°, g = 9.81: R = 20² × sin(90°) ÷ 9.81 = 400 ÷ 9.81 ≈ 40.78 m. The common slip is using sinθ instead of sin(2θ).