Example 1
Reference case: 20 m/s at 45° on level ground, with no air resistance. This is the farthest flight at that speed.
Z = 40.77 m
Enter the launch speed v₀ in meters per second and the angle α in degrees. The calculator finds the range on level ground from Z = v₀²·sin(2α)/g and also shows the peak height, the flight time, and the velocity components. At 20 m/s and 45° the range is 40.77 m: the farthest flight at that speed when launch and landing share one level. The model does not include a launch height h₀ or air resistance.
Enter the launch speed and the angle. Range, peak height, flight time, and the chart appear here.
For the peak height as a separate result, use the projectile height calculator. When the launch point is higher than the landing point, use the full projectile motion calculator.
Projectile range is the horizontal distance from the launch point to the landing point at the same height. When the launch speed v₀ makes an angle α with the horizontal, g is constant, and there is no air resistance, the range is Z = v₀² sin(2α)/g. That expression is the trigonometric form of the parabola Galileo described in 1638 as uniform horizontal motion combined with uniformly accelerated vertical motion (Halliday, Resnick, and Walker; Young and Freedman).
The calculator uses three numbers: speed v₀, angle α in degrees, and acceleration g. An empty g means 9.81 m/s². At 20 m/s and 45° the sine of the doubled angle is 1, so Z = 400 / 9.81 = 40.77 m. At 30° and at 60° sin(2α) has the same value, so both angles give 35.31 m.
What the result panel shows. Besides the range, the calculator reports the peak height H_max = v₀² sin²(α)/(2g), the flight time t = 2 v₀ sin(α)/g, the horizontal component v₀ cos(α), the vertical component v₀ sin(α), the range at 45° for the same speed, and the complementary angle 90° − α, which gives the same Z. At 20 m/s and 45° the peak is 10.19 m, the flight time is 2.88 s, and both components are 14.14 m/s. The bar chart compares the range at 15°, 30°, 45°, 60°, and 75° for the speed you entered.
Limits of the model. The formula requires launch and landing at one height. A flight from a tower is handled by the full projectile calculator, and a horizontal throw from an edge has its own calculator, because there the angle is 0° and the starting height is greater than zero. Air resistance is not in the arithmetic: in air, 45° is usually no longer the angle of greatest range. An angle of 0° and an angle of 90° both give Z = 0 m. So does zero speed.
The unit switch in the header converts meters to feet and meters per second to feet per second. The angle stays in degrees. Figures in the text stay in SI units as the reference. Choosing an example enters the same numbers into the fields and refreshes the panel and the chart.
Z = v02 sin(2α) / g
Hmax = v02 sin2(α) / (2g); t = 2 v0 sin(α) / g; vx = v0 cos(α); vy = v0 sin(α)
Model: level-ground projectile range (Galileo’s parabolic trajectory, 1638; the double-angle sine form in mechanics textbooks). The greatest Z at a fixed v₀ is at α = 45°, because sin(2α) = 1 there. Order of computation: v₀² · sin(2α) / g, with the angle converted from degrees to radians, plus H_max, flight time, and both components from the same velocity split.
Range Z is the horizontal distance between the place where the object leaves a level and the place where it returns to that level. The launch speed v₀ is inclined at an angle α to the horizontal. In the classroom model the horizontal acceleration is zero and the vertical acceleration is constant and equal to −g. The path is then a parabola, and the range on one level has the compact form Z = v₀² sin(2α)/g.
The peak height H_max = v₀² sin²(α)/(2g) describes the same flight but a different quantity: how high the object rises, not how far it lands. The flight time t = 2 v₀ sin(α)/g is the time from launch until the return to the launch level. The calculator reports Z as the main result, with H_max and t beside it.
Niccolò Tartaglia, in Nova Scientia (1537), described cannon shots and noted that an intermediate elevation, near 45°, reaches farther than a very flat or a very steep shot. Galileo gave the mathematical shape of the path in Discorsi e dimostrazioni matematiche intorno a due nuove scienze (1638), on the Fourth Day: horizontal motion is uniform, vertical motion is uniformly accelerated, and the combination is a parabola.
The form Z = v₀² sin(2α)/g is a later trigonometric writing of that same result. Sine, cosine, and the constant g in today’s notation do not appear in Galileo in this form. Mechanics textbooks (Halliday, Resnick, and Walker; Young and Freedman) derive the formula from Newton’s second law at constant g and with zero drag. The SI defines the meter and the second (BIPM, SI Brochure); the formula itself is a classroom result, not a statutory standard.
Step 1. Split the launch speed: vₓ = v₀ cos α stays constant, and vᵧ = v₀ sin α decreases by g each second. Step 2. The vertical coordinate returns to zero when vᵧ − g t = −vᵧ, so t = 2 v₀ sin α / g. The other root, t = 0, is the launch instant. Step 3. Range is the horizontal distance in that time: Z = vₓ · t = v₀ cos α · 2 v₀ sin α / g = v₀² · 2 sin α cos α / g. The identity 2 sin α cos α = sin(2α) gives Z = v₀² sin(2α)/g.
Step 4. The sine reaches 1 at 2α = 90°, so α = 45° and Z_max = v₀²/g. The angle 90° − α has the same sin(2α), because sin(180° − 2α) = sin(2α). Step 5. For v₀ = 20 m/s, α = 45°, and g = 9.81 m/s²: sin(90°) = 1, Z = 400 / 9.81 = 40.77 m, H_max = 200 / 19.62 = 10.19 m, t = 2.88 s, and vₓ = vᵧ = 14.14 m/s. At 30°, sin(60°) ≈ 0.866, so Z = 35.31 m.
The formula is exact only when landing is at the launch height, g is constant, and the drag force is zero. A launch from a balcony, a tower, or a table edge has a different final height and needs a quadratic equation in h₀, which the full-projectile calculator solves. A horizontal throw is the case α = 0° with h₀ greater than zero: this formula would then give Z = 0, which is false at an edge.
Air resistance grows with speed and with the object’s area. A ball, a shuttlecock, and a bullet land closer than they would in a vacuum, and the angle of greatest range usually drops below 45°. The vacuum formula at 800 m/s and 45° gives 640,000 / 9.81 ≈ 65,240 m. A real bullet in the atmosphere travels much less far. The calculator does not adjust the result for drag, wind, or spin.
Two angles that add to 90° give the same range on level ground and a different shape of path. At 20 m/s, 30° gives a peak of 5.10 m and a time of 2.04 s, while 60° gives a peak of 15.29 m and a time of 3.53 s. The range in both cases is 35.31 m. The panel shows the complementary angle so that pair is visible without a second calculation.
Speed enters as a square. At 45°, 20 m/s gives 40.77 m and 40 m/s gives 163.10 m. g sits in the denominator: the same 20 m/s at 45° with g = 1.62 m/s², the surface gravity of the Moon, gives 246.91 m. That is still a vacuum model, not a description of flight in a thin atmosphere.
A stone thrown at 10 m/s and 45° lands after 10.19 m. A low throw at 12 m/s and 15° gives only 7.34 m, because sin(30°) = 0.5. A speed of 16.6 m/s at 45° gives 28.09 m, about the length of a basketball court (28 m), if drag is ignored. A kick at 28 m/s and 40° in a vacuum gives 78.70 m; in air the range of the ball is shorter. A steep throw at 15 m/s and 80° has a range of 7.84 m and a peak of 11.12 m.
At high speeds the vacuum formula quickly stops describing air. 40 m/s at 45° is 163.10 m with no drag. 800 m/s at 45° is about 65,240 m in a vacuum, while the range of a rifle bullet in the atmosphere is many times smaller. The same formula remains valid as the no-drag limit; it is not a ballistic table.
The definition, derivation, and history come from mechanics textbooks and historical sources: G. Galilei, Discorsi e dimostrazioni matematiche intorno a due nuove scienze, 1638, Fourth Day (parabolic motion); N. Tartaglia, Nova Scientia, 1537 (gun elevation); D. Halliday, R. Resnick, J. Walker, Fundamentals of Physics, the chapter on motion in two dimensions; H. D. Young and R. A. Freedman, University Physics, “Motion in Two or Three Dimensions” (projectile motion); BIPM, The International System of Units (SI), 9th edition (meter, second). This page is for educational purposes.
Z is the range on level ground, v₀ the launch speed, and α the angle from the horizontal. At 20 m/s and 45° the result is 40.77 m.
Reference case: 20 m/s at 45° on level ground, with no air resistance. This is the farthest flight at that speed.
Z = 40.77 m
The same launch at 20 m/s, angle 30°. The path is lower, and the range is shorter than at 45°.
Z = 35.31 m
The complementary angle 60° at 20 m/s. The range matches 30°, and the peak is higher: 15.29 m instead of 5.10 m.
Z = 35.31 m
A kick at 28 m/s and 40° in the no-drag model. In air the range of the ball is shorter, because the formula does not subtract drag.
Z = 78.70 m
A throw at 25 m/s and 35° in a vacuum, for example as a comparison with a javelin throw when drag is ignored. In air the range is different, because a javelin also experiences lift.
Z = 59.87 m
A short throw at 18 m/s and 40° on a field, computed with no air resistance.
Z = 32.53 m
The same launch at 20 m/s and 45°, but g = 1.62 m/s², the surface gravity of the Moon. The Moon has no atmosphere, so the no-drag model fits well.
Z = 246.91 m
A low throw at 12 m/s and 15°. sin(2α) = sin(30°) = 0.5, so the range is clearly shorter than at 45°.
Z = 7.34 m
Twice the speed: 40 m/s at 45°. The range grows fourfold compared with 20 m/s, because v₀ is squared.
Z = 163.10 m
A launch at 16.6 m/s and 45° gives a distance close to the length of a basketball court (28 m), in the no-drag model.
Z = 28.09 m
A steep throw at 15 m/s and 80°. Most of the speed points upward: the range is short, and the peak is 11.12 m.
Z = 7.84 m
A stone thrown at 10 m/s and 45°. This is a short flight on level ground, with no drag.
Z = 10.19 m
The calculator finds the range of a projectile on level ground from Z = v₀² sin(2α)/g. For v₀ = 20 m/s and an angle of 45° the result is 40.77 m. Beside it, the panel shows a peak height of 10.19 m, a flight time of 2.88 s, and the velocity components.
Enter the launch speed in meters per second and the angle in degrees from 0 to 90. Enter g in meters per second squared; an empty field means 9.81. The unit switch in the header converts speed and length to feet, but the angle stays in degrees.
On level ground and with no air resistance the greatest range is at 45°, because sin(2α) then equals 1. At 20 m/s that range is 40.77 m. In air, drag usually moves the best angle below 45°.
Because sin(2α) = sin(180° − 2α), so the angles α and 90° − α share the sine of the doubled angle. At 20 m/s, 30° and 60° each give 35.31 m. The path at 30° is lower, and the path at 60° reaches higher: 5.10 m versus 15.29 m.
No. This formula assumes the height difference is zero. A flight from a tower, a wall, or an edge is handled by the full projectile calculator or the horizontal-throw calculator, where the starting height enters.
No. The model is the classroom one: constant g and no drag. Drag shortens the range of a ball, a shuttlecock, and a bullet, and the angle of greatest range is usually not 45°.
It is the peak of the same flight, H_max = v₀² sin²(α)/(2g), shown beside the range. At 20 m/s and 45° the peak is 10.19 m and the range is 40.77 m. A separate height calculator uses that peak as the main result.
Because the launch speed appears as a square. At 45°, 20 m/s gives 40.77 m and 40 m/s gives 163.10 m. The angle and g stay the same.
No. The calculator treats the number in the angle field as degrees. 45 means 45°, not π/4 written as 0.785. There are no radians in this field.
Both give a range of 0 m. At 0° there is no vertical component, so the object does not leave the level. At 90° there is no horizontal component: that is a vertical throw, and the landing is at the launch point.
An empty g field means 9.81 m/s², the gravitational acceleration used in classroom problems. Another value, for example 1.62 m/s² on the Moon, is entered directly. At 20 m/s and 45° that gives a range of 246.91 m.
Launch speed in this formula is non-negative, and the angle on level ground runs from 0° to 90°. Outside that interval the calculator shows a message instead of a range. Zero speed gives a result of 0 m, because the flight does not start.
Use Z = v₀²·sin(2α)/g. At 20 m/s and 45°, sin 90° = 1, so Z = 400/9.81 ≈ 40.77 m.
About 35.32 m. sin(60°) = √3/2, so Z = 400·√3/2 / 9.81 ≈ 35.32 m. The complementary 60° angle gives the same range.
Range is the horizontal distance back to launch level. Peak height is H = v₀²·sin²α/(2g). At 45°, H ≈ 10.19 m while Z ≈ 40.77 m.
No. Here launch and landing share one level. With h₀ ≠ 0 the flight time needs a square-root solution — that is the full projectile-motion model.
Because sin(2α) peaks at 1 when α = 45°. The condition is level ground, constant g, and no drag. From a height the optimal angle is usually smaller.
The definition, derivation, and history come from recognized physics sources.
Page updated November 3, 2026.