Projectile horizontal range:
how far a launch carries on level ground

Range as the product of two motions

A launched object does two jobs at once. Vertically it rises and falls under gravity; horizontally it keeps drifting forward at a steady speed, because with no air there is nothing to slow it. The landing point is simply the horizontal speed multiplied by the time it stays aloft.

Launch angle sets how the starting speed is split. A steeper angle gives more vertical speed, so a longer flight, but less horizontal speed. A shallow angle does the opposite. The best distance comes from a compromise between the two, and on level ground that compromise is exactly 45 degrees.

Everything below assumes launch and landing at the same height, no air resistance, and constant g. Those assumptions are what make a single clean formula possible.

The formula and the pieces behind it

Break the launch speed v into components: horizontal v × cos θ and vertical v × sin θ. Flight time on level ground is the time to go up and come back, 2 × v × sin θ ÷ g. Multiply horizontal speed by that time and the trigonometry collapses through the identity 2 sin θ cos θ = sin 2θ.

Worked example: 20 m/s at 45 degrees

A ball leaves a flat field at 20 m/s, angled 45° above the horizontal. Find where it lands, how long it flies and how high it goes.

Check with the horizontal speed: 20 × cos 45° is 14.14 m/s, and 14.14 × 2.88 s gives 40.7 m, which agrees with the formula. If you try this on a calculator, make sure it is set to degrees; radian mode returns nonsense for 45 and 90.

The formula depends on the square of speed. Throwing at 10 m/s from the same angle gives a quarter of the distance, about 10.19 m, not half.

How the launch angle shifts the landing point

Complementary angles, those that add up to 90°, give the same range. A 30° shot and a 60° shot both land at 35.31 m; the 60° one just takes longer and climbs higher, 15.3 m against 5.1 m. That is why a lofted throw and a flat throw can reach the same target.

The curve is quite flat near 45°. Being five degrees off costs less than one percent of the range, so exact angles matter less than most people expect.

When the simple range formula misleads

• Different heights. Launching from a cliff or a raised platform increases range, and the best angle falls below 45°. The simple formula underestimates, so use the full quadratic solution for flight time.
• Air resistance. Fast, light or wide objects, such as a shuttlecock or a football kicked hard, travel far shorter distances and take a steeper descent than the ideal arc.
• Spin and wind. Backspin can lift a golf ball well beyond the vacuum path, while a headwind shortens it.
• Low-g or high-altitude places. Replace 9.81 with the local value; the range scales as 1 ÷ g.

Working backwards: the speed needed to reach a target

Often the range is known and the speed is what you want. At 45°, sin(2θ) equals 1, so the formula reduces to R = v² ÷ g, and the required speed is v = √(R × g).

Speed grows only as the square root of range, so reaching a target five times as far needs only about 2.2 times the speed. That is why a gentle underarm toss and a hard overarm throw differ less in effort than their distances suggest, at least in the ideal model.

The same arithmetic on the Moon, where g is 1.62 m/s², turns the 20 m/s launch from the worked example into a range of 400 ÷ 1.62 ≈ 246.9 m, about six times the Earth figure. Everything in the equation stays the same; only g moves.

For anything off-level, solve the vertical motion for the flight time first and then multiply by the horizontal speed v × cos θ.

What a raised launch point does

Launching from a height h above the landing level adds to the range, because the object keeps travelling forward while it falls the extra distance. For a 20 m/s launch at 45° from a 2 m platform the flight lasts about 3.0 s rather than 2.88 s, which pushes the range from 40.8 m to roughly 42.7 m.

The best angle also drops below 45°. For a shot put released at around 2 m above ground at 13 m/s, the optimum angle sits near 42°, and athletes tend to throw even lower, typically in the mid-30s, because of how the body generates speed.

If your launch and landing heights differ by more than a few percent of the range, the neat sin 2θ formula becomes a lower bound, and it is better to solve for time of flight from the vertical equation.

Quick ways to check your answer

1. 1Estimate v² ÷ g first. This is the maximum range at 45°, so for 20 m/s it is about 40.8 m.
2. 2Multiply by sin(2θ). It cannot exceed 1, so the result must never be bigger than the estimate from step 1.
3. 3Compare complementary angles. If your numbers for 30° and 60° differ, an input is wrong.
4. 4Check units. Speed must be in m/s; if you start with km/h, divide by 3.6 before squaring.

A general physics calculator can be used to cross-check intermediate values such as v², while the final answer should always be compared against the v² ÷ g ceiling.