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How to calculate the range of a projectile (with a worked example)

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Split the launch speed into sideways and upward parts, find the time in the air, then multiply by the sideways speed. Formula, worked example and quick checks.

For a ball launched from flat ground with no air, the range is R = v² sin 2θ ÷ g, where v is the launch speed, θ the angle above the horizontal and g = 9.8 m/s². The projectile launcher works this out as you move its sliders, draws the path to scale and shows each step, including launches from a height where the short formula no longer applies.

What you need

  • The launch speed v, in metres per second (m/s).
  • The launch angle θ, measured up from the horizontal, in degrees.
  • The launch height h, in metres, if the ball leaves from above the landing level (a cliff, a table, a tall thrower). Use 0 for flat ground.
  • A calculator in degree mode, or just the tool.

Step by step

The key idea is that the sideways motion and the up-and-down motion happen at the same time but do not affect each other. Gravity only pulls down, so the sideways speed stays the same for the whole flight.

  1. Split the speed. Sideways: ux = v cos θ. Upward: uy = v sin θ. (Splitting a vector into parts is shown step by step in the vector adder.)
  2. Find the time in the air. From the ground, the ball rises and falls back to zero height, so T = 2 uy ÷ g. From a height h, solve h + uy t − ½ g t² = 0 for the positive t, which gives t = (uy + √(uy² + 2gh)) ÷ g.
  3. Multiply. Range R = ux × T. From the ground this tidies up to R = v² sin 2θ ÷ g, because 2 sin θ cos θ = sin 2θ.

With the tool:

  1. Leave the level on Class 11: no air.
  2. Set Launch speed v (5 to 40 m/s), Launch angle θ (0° to 90°) and Launch height h (0 to 30 m) with the sliders.
  3. Read Range, Highest point, Time of flight, Time to the top, Landing speed and Best angle for distance under The flight. The lines below them repeat your numbers inside each formula.
  4. Press Launch if you want to watch the ball fly along the drawn path.

A worked example

A footballer kicks a ball off flat ground at 25 m/s, 40° above the horizontal. Ignore air. How far does it go?

  • ux = 25 × cos 40° = 19.15 m/s; uy = 25 × sin 40° = 16.07 m/s.
  • Time of flight: T = 2 × 16.07 ÷ 9.8 = 3.28 s.
  • Range: R = 19.15 × 3.28 ≈ 62.8 m. The one-line formula agrees: 25² × sin 80° ÷ 9.8 = 625 × 0.9848 ÷ 9.8 = 62.81 m.
  • Highest point: H = uy² ÷ 2g = 16.07² ÷ 19.6 = 13.18 m.

On the page, a dashed curve also appears: the 50° kick at the same speed. It lands at the same 62.81 m, since sin 100° equals sin 80°, but it climbs to 18.71 m and stays up for 3.91 s. The best angle from the ground is 45°, which would reach 63.78 m.

Now the same kick from the top of a 5 m bank. Raise the height slider to 5 m. The time in the air grows to 3.57 s, so the range becomes 19.15 × 3.566 ≈ 68.29 m, and the best angle shown drops to 43°. A raised launch rewards a slightly flatter throw, because the ball gets extra falling time to keep travelling forward.

Check your answer

  • Pairs of angles. From flat ground, θ and 90° − θ give the same range. If your answers for 30° and 60° differ, look for a slip.
  • Ceiling. No range from flat ground can be more than v² ÷ g (the 45° result). For 25 m/s that is 625 ÷ 9.8 = 63.8 m, so 62.8 m at 40° is just under it, as it should be.
  • Half time at the top. From the ground, the time to the highest point is exactly half the time of flight (1.64 s of 3.28 s here).
  • Landing speed. With no air and the same launch and landing level, the ball lands at the speed it was launched (25 m/s here). Higher launches land faster.
  • Air makes it shorter. Real balls fall short of these numbers. Switch the tool to With air and pick a ball to see by how much; projectile drag goes further with wind.

Questions people ask

Why does 45° give the greatest range? The range follows sin 2θ, and a sine is largest (1) at 90°, so 2θ = 90° and θ = 45°. This holds only for flat ground with no air.

Do I use 9.8 or 10 for g? Use what your teacher or exam asks for. The tool uses 9.8 m/s². With 10 the answers come out about 2% smaller.

Can I use the equations of motion instead? Yes. The vertical part is a straight-line motion with acceleration −g, so s = ut + ½at² gives the time and the sideways part gives the distance. The equations of motion tool solves those one-line problems.

Sources

  • Projectile motion with constant gravity and no air: R = v² sin 2θ / g, T = 2v sin θ / g, H = v² sin²θ / 2g, and the launch-from-height case; standard school physics. g = 9.8 m/s².

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Tools in this article

Projectile launcher #L194 Launch a ball at any speed, angle and height and watch its path: the range, highest point, time of flight and landing speed worked out with your numbers, the matching angle that lands in the same place, the best angle for distance, and what air drag does to it. Equations of motion #L9 Drag a velocity-time graph and watch the rectangle and triangle under it add up to the displacement; see v = u + at, s = ut + ½at² and v² = u² + 2as with your numbers; solve any problem from three known values; and find a car's stopping distance. Vector adder #L193 Drag two arrows and watch their resultant drawn by the triangle law, the parallelogram law or by components, with its size and direction worked out. Then hang a weight from two strings and see Lami's theorem give each tension. Projectile drag #T636 Launch a ball at any speed, angle and height and see its real path through the air: quadratic drag from its size, mass and drag coefficient, a head- or tailwind, the range, highest point, time of flight, landing speed, the best angle for distance and how much the air costs compared with a vacuum.

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