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Projectile Motion Calculator

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Launch speed and angle in, trajectory out: range, peak height, flight time, and the speed and angle it lands at, with the arc drawn from the same points the numbers come from. It handles a launch height above the landing plane, which is the case the textbook 45° rule quietly assumes away — from a height the best angle is always less than 45°, and the tool works out how much less.

m/s
°

Above the horizontal. Negative throws downward.

m

Above where it lands. Zero for level ground.

How to use the projectile motion calculator

  1. 1Enter the launch speed in metres per second and the angle above horizontal.
  2. 2Set a launch height if the projectile does not start at ground level.
  3. 3Change gravity if you are not on Earth.
  4. 4Read the range and peak, and compare your angle against the optimum it reports.

Examples

The textbook case

Input
20 m/s at 45°, from ground level
Result
40.79 m range, 10.20 m peak, 2.88 s in the air

45° is optimal here, and only here — level ground is the condition.

Thrown from a height

Input
20 m/s at 45°, from 10 m up
Result
Range 53.5 m, and the optimum angle drops to about 35°

Starting high means hang time is already paid for, so a flatter throw goes further.

The same throw on the Moon

Input
20 m/s at 45°, lunar gravity
Result
246.9 m — six times the range, for the same effort

About the projectile motion calculator

Two problems, not one

The reason projectile motion is taught early is that it separates. Horizontally there is no force, so the motion is constant speed. Vertically there is a constant force, so the motion is constant acceleration. Neither half knows about the other; the only thing they share is the clock.

That is why a bullet fired horizontally and a bullet dropped from the same height hit the ground at the same moment — a demonstration that surprises people every time. The fired bullet's horizontal speed does nothing to delay its fall, because falling is a vertical problem and horizontal velocity is not part of it.

What air does to all of this

Drag opposes motion and grows with the square of speed, which changes the problem qualitatively rather than quantitatively. The trajectory becomes asymmetric: the projectile rises further than it falls in horizontal distance, and comes down more steeply than it went up. The optimum angle drops well below 45° — for a golf ball it is nearer 30° once lift from backspin is included.

There is no closed-form solution for the general case, which is why this calculator does not pretend to include one. Real ballistics is solved by numerical integration over small time steps, with drag coefficients measured in a wind tunnel for the specific projectile. The vacuum answer is still worth having: it is the upper bound the real range sits under, and it is exactly right about the things air does not touch, such as which of two launches spends longer in the air.

Gravity is a parameter, not a constant

Earth's 9.81 m/s² is itself an average: gravity is about 0.5% stronger at the poles than at the equator, because the Earth bulges and because the equator is moving. That variation is far too small to notice in a thrown ball and large enough to matter to a long-range artillery table.

Changing worlds changes everything. Lunar gravity is a sixth of Earth's, and since range scales inversely with g, the same throw goes six times as far and stays in the air six times as long. It is the clearest demonstration that range is not a property of the throw alone — the same arm produces a wildly different result depending on what is pulling the projectile down.

Frequently asked questions

Why is 45° the best angle, and when is it not?
Range is horizontal speed multiplied by time in the air. Raising the angle buys hang time and costs horizontal speed, and on level ground those trade off exactly evenly at 45°. The condition is level ground. Launch from a height and some hang time is free, so the optimum shifts lower — from 10 m up at 20 m/s it is about 35°, and the higher you start the flatter the best throw becomes.
Does this include air resistance?
No, and that matters more than it sounds. This is the vacuum trajectory. For a dense, slow, small object over a short distance it is close: a thrown stone behaves roughly like this. For anything light or fast it is badly wrong — a struck golf ball travels around half the range predicted here, and a bullet a small fraction of it. Drag rises with the square of speed, so it is not a correction factor you can apply afterwards; the path stops being a parabola at all.
Why do the two halves of the arc look symmetrical?
From level ground they are, exactly. The projectile takes as long coming down as going up and lands at the same speed it left, mirrored below the horizontal. That symmetry is a consequence of nothing acting horizontally and gravity acting constantly, and it is the first thing air resistance destroys — a real trajectory falls more steeply than it rose.
What happens to horizontal speed during the flight?
Nothing at all, in this model. Gravity pulls straight down, so it changes only the vertical component; the horizontal speed the projectile leaves with is the horizontal speed it lands with. That is why the range is simply horizontal speed times flight time, and why the whole problem separates into two independent one-dimensional ones — which is the actual lesson behind the exercise.
Can I use it for a ballistic or sporting calculation?
For understanding the shape of the problem, yes. For a real shot, no. Anything where the answer matters — ballistics, artillery, competitive sport — needs drag, spin, wind and often air density and Coriolis, and those are modelled numerically rather than in closed form. Treat this as the frictionless ideal the real answer sits below.

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