Projectile Range Calculator
From launch speed and angle, get the range, maximum height, and flight time over level ground.
Max height: 10.19 m · Flight time: 2.88 s · at 45° the range is greatest for a fixed speed
Height above the ground versus horizontal distance — a symmetric parabola peaking at the midpoint of the range.
On level ground the range is R = v²·sin(2θ) ÷ g. Launching at v = 20 m/s and θ = 45° with g = 9.81 m/s² gives R = 20² × sin(90°) ÷ 9.81 = 400 ÷ 9.81 = 40.77 m. A 45° angle produces the maximum range for any fixed launch speed.
What projectile range means
The range is the horizontal distance a projectile covers before returning to its launch height. It depends only on the launch speed, the launch angle, and gravity — because horizontal and vertical motion are independent. The horizontal speed stays constant while gravity pulls the vertical speed to zero at the peak and back down, and the two combine into the familiar arc. This calculator assumes flat, level ground and no air resistance; for the full velocity-component breakdown see the projectile motion calculator.
v = launch speed, θ = launch angle, g = 9.81 m/s² (level ground)
Worked example
A ball launched at v = 20 m/s, θ = 45°, on Earth (g = 9.81 m/s²):
- 1 Convert the angle to radians. θ = 45° = 0.7854 rad, so sin(2θ) = sin(90°) = 1 and sinθ = 0.7071.
- 2 Apply the range formula. R = v² × sin(2θ) ÷ g = 20² × 1 ÷ 9.81 = 400 ÷ 9.81 = 40.77 m.
- 3 Find the maximum height. H = v² × sin²θ ÷ (2g) = 400 × 0.5 ÷ 19.62 = 200 ÷ 19.62 = 10.19 m.
- 4 Find the flight time. T = 2 × v × sinθ ÷ g = 2 × 20 × 0.7071 ÷ 9.81 = 28.28 ÷ 9.81 = 2.88 s.
Range vs launch angle (as a fraction of the 45° maximum)
Range scales with sin(2θ), so complementary angles that add to 90° — like 30° and 60° — share the same range. 45° is the peak.
| Angle θ | sin(2θ) | Range fraction of max | Range at v = 20 m/s |
|---|---|---|---|
| 15° | 0.500 | 50.0% | 20.39 m |
| 30° | 0.866 | 86.6% | 35.31 m |
| 45° | 1.000 | 100% | 40.77 m |
| 60° | 0.866 | 86.6% | 35.31 m |
| 75° | 0.500 | 50.0% | 20.39 m |
Assumptions and the 45° optimum
This is the drag-free, level-ground model. Air resistance is ignored — standard for introductory physics but the reason real balls fall short of these numbers — and the projectile is assumed to land at the same height it left from. Launching from a cliff or a raised platform lengthens the flight and needs the fuller kinematic equations.
Why 45° wins: range depends on sin(2θ), which peaks at sin(90°) = 1 when θ = 45°. Steeper angles trade horizontal reach for height, and shallower ones give too little hang time. Maximum height, by contrast, keeps climbing all the way to a straight-up 90° throw — which has zero range.