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Projectile Launch Angle Calculator

Enter range and speed to find the required launch angle.

How it is calculated

For a projectile that lands at the height it was launched from, with no air resistance, the range is R = v² × sin(2θ) ÷ g. The tool solves it backwards: sin(2θ) = R × g ÷ v², where R is the target distance in metres, v the launch speed in m/s and g gravity in m/s². The low angle is θ = ½ × arcsin(sin 2θ) and the high angle is (180° − arcsin(sin 2θ)) ÷ 2; the two always add up to 90°. If g is left empty or is not above zero, the tool uses 9.81. A target beyond the maximum range v² ÷ g makes sin(2θ) larger than 1, and then there is no result. Speed must be above zero and range cannot be negative. The angles are in degrees.

Worked example

You want to hit a target 60 m away on level ground with a launch speed of 30 m/s and g = 9.81. First sin(2θ) = 60 × 9.81 ÷ 30² = 588.6 ÷ 900 = 0.6540. The arcsine of 0.6540 is 40.84°, so the low angle is half of that, 20.422°. The high angle is (180 − 40.84) ÷ 2 = 69.578°. Check that they add up to 90°: 20.422 + 69.578 = 90.000. Both angles land at exactly 60 m in this idealised model, the flat one arriving sooner and the lofted one later and more steeply.

Try this example →

Common mistakes

  • Entering the range beyond what the speed allows: the maximum is v² ÷ g, about 91.74 m at 30 m/s, and anything further returns no result.
  • Using km/h or feet per second for the speed while the range is in metres; speed must be in m/s.
  • Expecting the real-world trajectory to match; the model ignores air resistance and the height of the launch point above the target.
  • Treating the two angles as alternatives that differ in range; both hit the same range, they differ only in flight time and steepness.
  • Overlooking that the target at the maximum range has a single angle of 45°, where the low and high angles coincide.

About this calculator

The projectile launch angle calculator solves the classic ballistics problem in reverse: given a target range R and a launch speed v on level ground, it finds the angle θ using θ = ½ arcsin(gR / v²), where g is gravitational acceleration. Because sin(2θ) = sin(180° − 2θ), this equation always has two valid solutions — a low, flat trajectory and a high, lofted one — that hit the exact same range.

The low angle gets there faster and flatter (favored when speed matters, or when the flight path must clear low obstacles), while the high angle spends longer in the air and comes down more steeply. Physics students verifying textbook kinematics problems, and hobbyists in fields like archery, ballistics, and sports science (javelin, shot put, artillery) use this calculator to find both possible launch angles instead of solving the trig by hand.

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Comments (1)

  • Editorial team

    Watch the sin(2θ) line as it creeps toward 1. At 30 m/s, a target 90 m away reads 0.9810 and gives 39.407° and 50.593°, but at 29 m/s the same target is out of reach and nothing is returned. Near the limit the two angles bunch together and a small speed error matters a lot, so keep a margin if the speed is only an estimate. Is your launch speed measured or guessed in your problem?

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