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Projectile motion calculator

Enter the launch speed, angle and starting height to get the range, time of flight, peak height and impact speed, with the path drawn. Metric or US units.

Private: calculated on your device. Nothing you type is sent anywhere.

Next: Moving at a steady speed instead? Use the speed, distance and time calculator. Breaking a velocity into parts is right-triangle math: the triangle calculator shows it.

What this calculator does

It solves the classic physics problem of something thrown, kicked or hit near the ground. Give it the launch speed, the angle above the horizontal and the starting height, and it works out how far the object goes, how long it's in the air, how high it gets and how fast it lands. Use it to check homework or to compare angles for a real throw.

It assumes there's no air resistance and that gravity is the only force, the textbook model (OpenStax calls this neglecting air resistance; NASA calls it ballistic flight). Gravity is the standard value g = 9.80665 m/s², which NIST lists as exact. In US units that's 9.80665 ÷ 0.3048 ≈ 32.174 ft/s².

How it's calculated

Split the launch speed v₀ at angle θ into a sideways part and an upward part. The sideways speed never changes; the upward speed falls by g every second.

vx = v₀ cos θ   vy = v₀ sin θ
height: y(t) = h₀ + vyt − ½gt²
time of flight: t = (vy + √(vy² + 2gh₀)) ÷ g
range = vx × t   max height = h₀ + vy² ÷ 2g at t = vy ÷ g

The time of flight is the positive root of y(t) = 0. When h₀ = 0 it reduces to T = 2v₀ sin θ ÷ g and range R = v₀² sin 2θ ÷ g. At impact the velocity is vx sideways and vy − gt downward.

Worked example

A ball leaves the ground at 20 m/s at 45°. Then vx = vy = 20 × 0.70711 = 14.142 m/s.

  • Time of flight: 2 × 14.142 ÷ 9.80665 = 2.8842 s
  • Range: 20² × sin 90° ÷ 9.80665 = 400 ÷ 9.80665 = 40.789 m
  • Maximum height: 14.142² ÷ (2 × 9.80665) = 10.197 m, reached after 1.4421 s
  • Impact: 20 m/s at 45° below the horizontal, the mirror image of the launch

Throw it from a 20 m cliff instead, horizontally at 15 m/s, and it falls for √(2 × 20 ÷ 9.80665) = 2.0196 s, landing 30.294 m out at 24.845 m/s.

Angle, range and height

Angle (20 m/s, from the ground)RangeMax height
15° or 75°20.39 m15°: 1.37 m · 75°: 19.03 m
30° or 60°35.32 m30°: 5.10 m · 60°: 15.30 m
45°40.79 m (the most)10.20 m

On level ground, angles that add up to 90° give the same range, because sin 2θ is the same for both. From a raised starting point the best angle is a little under 45°; the comparison table in the result shows this for your numbers.

When the answer doesn't apply

  • Air resistance. Real balls, arrows and bullets slow down, so they fall short of these ranges, often by a lot at high speeds. OpenStax notes that with air resistance the best angle is somewhat below 45°.
  • Very long or very high shots. Gravity is taken as constant and the ground as flat, which stops being true over many kilometers.

Sources

How we checked this

Before every update, this calculator is run in a real browser with these known answers, and it must match all of them:

  • 20 m/s at 45° from the ground: range 400 ÷ 9.80665 = 40.7886 m ✓
  • 20 m/s at 45°: time of flight 2.88419 s ✓
  • US units: 100 ft/s at 30° from 6 ft lands 279.188 ft away ✓
  • 60 mph at 35° from 3 ft hits the ground at 60.7432 mph ✓
  • Thrown flat at 15 m/s from a 20 m cliff: lands 30.2943 m out ✓
  • Edge: straight up at 10 m/s peaks at 5.09858 m ✓
  • Edge: 0° from the ground lands straight away ✓

Last checked October 26, 2026. Spotted something wrong? The math is done in your browser with the formula shown on this page, so you can check any result by hand.

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Questions people ask

What angle gives the maximum range?

45° on level ground, with no air resistance. The range is v₀² sin 2θ ÷ g, and sin 2θ is largest (1) when θ = 45°. From a raised starting point the best angle is a little less than 45°.

How do you find the time of flight of a projectile?

From level ground, T = 2v₀ sin θ ÷ g. From a starting height h₀, solve h₀ + v₀ sin θ · t − ½gt² = 0, which gives t = (v₀ sin θ + √((v₀ sin θ)² + 2gh₀)) ÷ g.

How do you find the maximum height?

The upward speed drops to zero at the top, after t = v₀ sin θ ÷ g. The height then is h₀ + (v₀ sin θ)² ÷ 2g. A launch at 20 m/s and 45° peaks at about 10.2 m.

Does the mass of the object matter?

Not in this model. Without air resistance every object falls with the same acceleration g, so a heavy and a light ball launched the same way follow the same path.

What value of g does it use?

The standard acceleration of gravity, 9.80665 m/s² (about 32.174 ft/s²). Real gravity varies a little from place to place.

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