Example 1
A ball on a field, the reference example: speed 20 m/s at 60°. The model has no air drag.
H_max = 15.29 m · Z = 35.31 m
Projectile Maximum Height Calculator
Enter the initial speed v₀ in meters per second and the angle α in degrees. The calculator finds the peak height above the launch level from H_max = v₀²·sin²(α)/(2g). A ball thrown at 20 m/s and 60° reaches 15.29 m, higher than the same 20 m/s at 45°, where the peak is 10.19 m. The model does not add a launch height h₀, does not compute range Z, and ignores air resistance.
Result
Enter the initial speed and the angle. Peak height, range, and the chart will appear here.
For the range on level ground, use the projectile range calculator. For a launch from a tower or another height, use the full projectile motion calculator.
How it works
v₀20 m/s
α60°
H_max15.29 m
In Galileo’s model, projectile motion is uniform horizontal motion combined with free fall. Peak height is H_max = v₀² sin²(α) / (2g). At 20 m/s and 60°, the sine of the angle is √3/2, so sin²(α) = 3/4. Then H_max = 400 · (3/4) / (2 · 9.81) = 300 / 19.62, which the panel rounds to 15.29 m. The same speed at 45° gives 10.19 m, at 30° gives 5.10 m, and at 90° gives 20.39 m.
The calculator uses three numbers: initial speed v₀ in m/s, angle α in degrees, and acceleration g in m/s². A blank g means 9.81. The angle must lie from 0° to 90°, and the speed cannot be negative. 60 in the field is 60°, not 60 radians. A comma and a period are the same number.
What the result panel shows. Besides H_max, the calculator reports range Z = v₀² sin(2α)/g, the components v_y = v₀ sin(α) and v_x = v₀ cos(α), climb time t = v_y/g, and flight time 2t. For the reference values that is 15.29 m, 35.31 m, 17.32 m/s, 10.00 m/s, 1.77 s, and 3.53 s. The chart draws H_max at 15°, 30°, 45°, 60°, 75°, and 90° for the entered speed.
Limits of the model. The formula is for a launch from ground level that lands on the same level, with no air drag. An angle of 0° gives H_max = 0, because there is no upward component; that is not a horizontal throw from a ledge, where the launch height is given. A tower launch is covered by the full-projectile calculator, with H_max = h₀ + v_y²/(2g). Level-ground range alone is covered by the range calculator. At 20 m/s and 45° the range is 40.77 m and the peak is 10.19 m.
The closer the angle gets to 90°, the higher the peak at the same v₀, while the range shrinks. 45° is a compromise for range, not for height. Speed is squared: 40 m/s at 60° gives 61.16 m, four times 15.29 m. A larger g cuts the peak. On the Moon, with g = 1.62 m/s², the same 20 m/s and 60° give 92.59 m.
The unit switch in the header changes meters to feet and meters per second to feet per second. The numbers in the text stay in SI as the reference. The button “Fill this example” enters that example’s numbers into the fields and refreshes the panel and the chart. The fields are empty when the page opens.
Formula
Hmax = v02 sin2(α) / (2g)
Z = v02 sin(2α) / g on level ground; t = v0 sin(α) / g, flight time = 2t.
Model: Galilean projectile motion (uniform horizontal motion combined with free fall; Galileo Galilei, Discorsi, 1638, Fourth Day). The calculator finds H_max, and beside it Z, the components, and the times. It does not compute air drag or a launch from a height h₀.
Theory, derivation, and scope
Definition of the maximum height of a projectile
The maximum height of a projectile is the vertical distance from the launch level up to the highest point of the path. The object leaves with speed v₀ at an angle α above the horizontal. The horizontal component v_x = v₀ cos(α) stays constant while there is no drag. The vertical component v_y = v₀ sin(α) falls under constant acceleration g and reaches zero at the apex. From there the object descends.
The formula this calculator uses is H_max = v₀² sin²(α)/(2g). That is the same relation as H_max = v_y²/(2g), because v_y = v₀ sin(α). In SI the result is in meters. At 20 m/s and 60°, the product v₀² sin²(α) is 300, the denominator 2g is 19.62, and the quotient rounded to two decimal places is 15.29 m. That pair of numbers comes back in the later examples. At 45° the same speed gives 10.19 m, and at 30° it gives 5.10 m.
Historical sketch: Galileo, Tartaglia, and the parabola
The split of a throw into horizontal and vertical motion comes from Galileo. In Discorsi e dimostrazioni matematiche intorno a due nuove scienze (Leiden, 1638), on the Fourth Day, he showed that the path of a projectile in the no-drag model is a parabola: horizontal motion is uniform, and vertical motion is uniformly accelerated. Galileo did not write the formula in today’s notation with a sine and with g in meters per second squared. The closed form H_max = v₀² sin²(α)/(2g) is the later textbook writing of that split.
Earlier, Niccolò Tartaglia in Nova Scientia (Venice, 1537) studied the firing angle and associated the greatest range on level ground with an elevation of 45°. In the vacuum model that conclusion follows from sin(2α), which equals 1 at 45°. Tartaglia did not derive the modern height formula. Today both results, range and peak, are in every textbook chapter on two-dimensional kinematics. The meter and the second used here are SI units described in the BIPM brochure.
Derivation of H_max = v₀² sin²(α)/(2g)
Step 1. The y axis points upward and the launch is at y = 0. Vertical acceleration is −g and horizontal acceleration is 0. The initial components are v_x = v₀ cos(α) and v_y0 = v₀ sin(α). Step 2. Vertical speed in time is v_y(t) = v_y0 − g t. At the apex v_y = 0, so the climb time is t = v_y0 / g. Step 3. Vertical position is y(t) = v_y0 t − ½ g t². Substituting t = v_y0 / g gives y = v_y0² / g − ½ v_y0² / g = v_y0² / (2g).
Step 4. Because v_y0 = v₀ sin(α), the peak height is H_max = v₀² sin²(α) / (2g). At 20 m/s and 60°: sin 60° = √3/2, sin² = 3/4, and 300 / 19.62 is 15.2905…, so the panel shows 15.29 m. The climb takes 1.77 s, and the flight time on level ground, back to y = 0, is twice that: 3.53 s. Range is then v_x times the flight time, which reduces to Z = v₀² sin(2α) / g = 35.31 m. The calculator treats H_max as the main result and Z as a companion value.
Assumptions: ground-level launch, constant g, no drag
The calculator assumes constant acceleration g, no air drag, a point mass, and a launch from ground level that also lands on that level. The angle is measured upward from the horizontal, from 0° to 90°. At 0° the vertical component vanishes and H_max = 0. At 90° the motion is a vertical throw, H_max = v₀²/(2g), and the range is zero. The g field may be left blank: the calculator then uses 9.81 m/s².
The formula has no launch height h₀. A throw from a tower, a roof, or a ledge has a different flight time and a different range, and the peak is h₀ + v_y²/(2g) when there is an upward component. That is the scope of the full-projectile calculator. Air drag lowers both range and peak; at high speed the path stops being a symmetric parabola. Wind, ball spin, and g that changes with height are also outside this model. The calculator does not add those corrections.
Height, range, climb time, and the velocity components
The panel reports H_max, range Z, the vertical component, the horizontal component, the climb time, and the flight time. At 20 m/s and 60° that is 15.29 m, 35.31 m, 17.32 m/s, 10.00 m/s, 1.77 s, and 3.53 s. The angles 30° and 60° share the range 35.31 m, because sin(2·30°) = sin(2·60°), while the heights differ: 5.10 m and 15.29 m. Range alone therefore does not determine the peak. A separate page computes Z only.
At the apex the vertical speed is zero and the speed equals v_x. Potential energy at that point depends on mass, which this calculator does not use, so it does not compute that energy. Doubling v₀ multiplies H_max by four: 40 m/s and 60° give 61.16 m. Doubling g cuts the height in half. Changing the angle at fixed v₀ moves the result along sin²(α), from zero at 0° to v₀²/(2g) at 90°.
Units, orders of magnitude, and throws around us
The meter is the SI unit of length, the second is the unit of time, and the meter per second is the unit of speed. Acceleration is in m/s². After the header is switched to US units, the panel shows feet and feet per second and, in parentheses, the SI value. The page text stays in meters. A catalog speed in km/h has to be divided by 3.6 before it is entered: 72 km/h is 20 m/s. In US customary units, miles per hour convert to feet per second before the same formula.
The magnitudes below illustrate the formula; they are not a measurement from a field. A short throw of 10 m/s at 45° reaches 2.55 m. A garden hose in the no-drag model, 8 m/s at 55°, reaches 2.19 m; a real stream sits lower. A lob of 12 m/s at 70° reaches 6.48 m. The reference, 20 m/s and 60°, is 15.29 m, under a typical 16 m hall clearance in the no-drag model. A hard clearance of 25 m/s at 50° reaches 18.69 m. On the Moon the same 20 m/s and 60° at g = 1.62 m/s² reach 92.59 m. A gym, a field, and the Moon differ here only by the numbers in the same formula.
Sources
The definition, the derivation, and the historical sketch come from physics textbooks and historical sources: Galileo Galilei, Discorsi e dimostrazioni matematiche intorno a due nuove scienze, Leiden 1638, Fourth Day; Niccolò Tartaglia, Nova Scientia, Venice 1537; D. Halliday, R. Resnick, and J. Walker, Fundamentals of Physics, two-dimensional kinematics; H. D. Young and R. A. Freedman, University Physics, the chapter on motion in two and three dimensions; BIPM, The International System of Units, 9th edition (the meter and the second). This page is for educational purposes and does not replace a measurement that includes air drag.
How to use
- Enter the initial speed v₀ in meters per second, for example 20. That is the speed at a ground-level launch.
- Enter the angle α in degrees, for example 60. The field takes degrees. 60 means 60°, not radians.
- Leave g blank for Earth, or enter another value, for example 1.62 on the Moon. At 20 m/s and 60° the calculator shows 15.29 m.
- Check the panel: range, velocity components, climb time, and flight time. At 0° the height is 0. At 90° and 20 m/s the height is 20.39 m.
- Clear removes the entries and hides the panel. The unit switch in the header changes meters to feet. A launch with a starting height is covered by the full-projectile calculator.
Symbol glossary
H_max is the peak height, v₀ the launch speed, α the angle above the horizontal, and g the acceleration. At 20 m/s and 60° the result is 15.29 m.
- H_max
- Peak height [m], H_max = v₀² sin²(α)/(2g). At 20 m/s and 60° it is 15.29 m.
- v₀
- Initial speed [m/s]. It is squared in the formula.
- α
- Angle above the horizontal [°]. Range 0 to 90. Not radians.
- g
- Acceleration [m/s²]. A blank field means 9.81.
- v_y
- Vertical component [m/s], v₀ sin(α). It falls to zero at the apex. For the reference values, 17.32 m/s.
- Z
- Level-ground range [m], v₀² sin(2α)/g. A companion value; a separate page computes only that.
Real-life examples
Example 2
The same throw at 45°: speed 20 m/s. The peak is lower than at 60°, and the level-ground range is at its largest.
H_max = 10.19 m · Z = 40.77 m
Example 3
A flat arc: 20 m/s at 30°. The range is the same as at 60°, because sin(60°) = sin(120°), while the peak is clearly lower.
H_max = 5.10 m · Z = 35.31 m
Example 4
A steep lob: 15 m/s at 80°. Almost the entire speed is vertical, so the range is short while the peak stays high.
H_max = 11.12 m · Z = 7.84 m
Example 5
A tennis lob: 12 m/s at 70°. In this vacuum model the ball clears the opponent and lands a short distance beyond the net.
H_max = 6.48 m · Z = 9.44 m
Example 6
A garden-hose stream in the school approximation: 8 m/s at 55°. Water in air has drag, so the real arc sits lower.
H_max = 2.19 m · Z = 6.13 m
Example 7
The same throw on the Moon: 20 m/s at 60° with g = 1.62 m/s². With the smaller acceleration the peak is several times higher.
H_max = 92.59 m · Z = 213.83 m
Example 8
A hard clearance: 25 m/s at 50°. This is still a ground-level launch with no drag; once air drag is included, the arc is lower.
H_max = 18.69 m · Z = 62.74 m
Example 9
Double the reference speed: 40 m/s at the same 60°. The peak is four times higher, because v₀ is squared in the formula.
H_max = 61.16 m · Z = 141.25 m
Example 10
A short textbook throw: 10 m/s at 45°. With sin²(45°) = 1/2, the height is v₀²/(4g).
H_max = 2.55 m · Z = 10.19 m
Common questions
What does this calculator compute, and from which formula?
The calculator finds the peak height of a projectile from H_max = v₀² sin²(α)/(2g) and, alongside it, the level-ground range Z = v₀² sin(2α)/g. At 20 m/s and 60° the peak is 15.29 m and the range is 35.31 m.
Which units should I enter for speed, angle, and g?
Speed in meters per second, the angle in degrees, and g in meters per second squared. The result is in meters. A blank g field means 9.81 m/s². After the header is switched to US units, speed is in feet per second, g is in ft/s², and height is in feet; the panel also shows the SI value.
Why does the peak rise with the angle while range is largest at 45°?
Height depends on sin²(α), so at the same v₀ it rises from 0° to 90°. Range on level ground depends on sin(2α) and is largest at 45°. At 20 m/s the peak at 30°, 45°, and 60° is 5.10 m, 10.19 m, and 15.29 m, and the range at 45° is 40.77 m.
What do 0° and 90° give?
At 0° the vertical component vanishes, so H_max = 0 and the range is zero as well. At 90° the formula becomes a vertical throw: H_max = v₀²/(2g). At 20 m/s that is 20.39 m, and the range drops to zero. That is not a horizontal throw from a tower: there the launch height is given, while this calculator starts from ground level.
Does the calculator include air drag or a launch from a height?
No. The model is a vacuum, constant g, and a launch from ground level that also lands on that level. In air the peak sits lower than this formula. A launch from a tower, with a nonzero starting height, is covered by the full-projectile calculator: there H_max = h₀ + v_y²/(2g).
How is height on this page different from range?
The main result is H_max, the vertical distance from the launch level up to the apex. Range Z appears on the panel as a companion value and has its own calculator. At 20 m/s and 45° the range is 40.77 m and the peak is 10.19 m.
Why does doubling the speed give four times the height?
Because speed enters the formula as a square. At 60° and g = 9.81 m/s², 20 m/s gives 15.29 m and 40 m/s gives 61.16 m. The angle and the acceleration stay the same.
How long does the climb to the apex take?
The climb time is t = v₀ sin(α)/g, and the flight time on level ground is 2t. At 20 m/s and 60° the vertical component is 17.32 m/s, the climb takes 1.77 s, and the full flight takes 3.53 s. At the apex the vertical speed is zero; only the horizontal component remains.
Are a comma and a period equivalent, and are the fields filled in advance?
Yes, a comma and a period are the same number: the calculator reads 15,5 and 15.5 as the same speed. The fields are empty when the page opens. The hint in a field is not an entered value.
What does the panel show besides the height?
Besides H_max, the panel shows range Z, the components v_x and v_y, the climb time, and the flight time. Under the numbers is a sentence that substitutes those values into the formula. The bar chart shows H_max at 15°, 30°, 45°, 60°, 75°, and 90° for the entered speed and the same g.
Why is there no result outside the angle range or when g is zero?
The angle must lie between 0° and 90° inclusive. Initial speed cannot be negative. Acceleration g must be greater than zero, because it sits in the denominator. At g = 0 the height formula is undefined.
How do I calculate the maximum height of a projectile?
Use H = v₀²·sin²α/(2g). At 20 m/s and 60°, sin 60° = √3/2, so H ≈ 15.29 m above the launch level.
What is H at 20 m/s and 30°?
About 5.10 m. sin 30° = 1/2, so sin²α = 1/4 and H = 100/(2·9.81) ≈ 5.10 m.
How is height different from projectile range?
Height is the peak above launch: H = v₀²·sin²α/(2g). Level-ground range is Z = v₀²·sin(2α)/g and peaks at 45°. Range has its own page.
Does the result include a launch height h₀?
No. The primary result is the climb above the launch level. Height above the ground is h₀ + H; a full path with h₀ ≠ 0 belongs on the full projectile-motion page.
Does mass change the peak height?
No. In the ideal model without drag, H depends only on v₀, α, and g. Mass cancels out of the vertical motion equations.
Sources
The definition, the unit, the history, and the derivation come from established physics and metrology sources.
- Wikipedia — Projectile motion
- Wikipedia PL — rzut ukośny
- BIPM — The International System of Units (SI), 9th edition
- NIST SP 330 — The International System of Units
- Galileo Galilei, Discorsi e dimostrazioni matematiche intorno a due nuove scienze, Leiden 1638, Fourth Day.
- Tartaglia, N., Nova Scientia, Venice 1537.
- Halliday, D.; Resnick, R.; Walker, J., Fundamentals of Physics — two-dimensional kinematics.
- Young, H. D.; Freedman, R. A., University Physics — motion in two and three dimensions.
Page updated November 3, 2026.