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You throw straight up, drop from an edge, or kick at an angle. Below are three models in one table: when to compute H_max, when Z, and when the full path from a launch height.
Everything in one tool: full projectile motion. Step-by-step theory: Physics without mysteries.
On this page: v₀ is initial speed, α is the launch angle, h₀ is launch height. H_max is the greatest height, Z is the range, t is flight time. g is gravitational acceleration, about 9.81 m/s².
You throw a ball straight up, drop a stone from a bridge, or kick a ball at an angle. In the simple model the same physics applies: constant g downward and no air drag. Only the launch conditions change, and those decide which formula and calculator you need.
Here a “throw” means motion under constant gravity. Acceleration has only a vertical component of magnitude g = 9.81 m/s² downward. Horizontally, if no other forces act, speed stays constant. That shared base supports the three models below.
The three models are not “different physics”. Vertical throw: v0 only vertical. Horizontal throw: launch from height h with v0 only horizontal. Angled throw: v0 at angle α, so you immediately have vx = v0 cos α and vy = v0 sin α.
Shared intuition: x and y motions are independent. In x: uniform (or rest). In y: uniformly accelerated with a = -g if the axis points up. The whole table at the bottom of the page is built from those blocks.
Vertical throw upward: at the top v = 0, so from v² = v0² - 2gH you get:
H_max = v0² / (2g)
Rise time is t_up = v0/g. Return to the same height takes the same time, so the full up-and-down flight is 2·t_up = 2·v0/g. There is no horizontal range here: the x-axis is idle.
A mini numeric example used in all three stories: v0 = 10 m/s, g = 9.81 m/s². Vertically, H_max ≈ 5.1 m. Rise time ≈ 1.02 s, and the full up-and-back flight ≈ 2.04 s. The same numbers appear in the problem below.
A horizontal throw leaves an edge, roof or bridge. Flight time comes from the vertical fall: with vy = 0 at launch, h = ½·g·t² gives:
t = √(2h/g)
Horizontal range is Z = v0·t because vx = v0 = const. The highest point above ground is usually the launch itself: there is no rise “above the roof”. For comparison: with v0 = 10 m/s and h ≈ 5.1 m, flight time is √(2·5.1/9.81) ≈ 1.02 s and Z ≈ 10.2 m. One speed, another story than the pure vertical case.
A second micro-check locks the order of magnitude. From a roof h = 20 m at v0 = 8 m/s, flight time ≈ √(2·20/9.81) ≈ 2.02 s, so Z ≈ 16.2 m. If your head says “hundreds of meters” for a calm roof toss, you almost certainly mixed speed units.
An angled throw on flat ground (launch and landing at the same height) has the classic formulas:
Z = v0² · sin(2α) / g
H_max = v0² · sin²α / (2g)
Flight time is 2·v0·sin α / g. With no drag, maximum range sits at α = 45°. With the same v0 = 10 m/s at 45° you get Z = 100/9.81 ≈ 10.2 m and H_max = 50/19.62 ≈ 2.55 m. Compare with vertical (H_max ≈ 5.1 m, Z = 0) and with horizontal from h ≈ 5.1 m (Z ≈ 10.2 m, no rise above launch). The table below keeps those differences in one place.
From a tower (h0 > 0) the “flat” Z formula is not enough. You must solve vertical motion to y = 0 with nonzero h0, get flight time, then Z = vx·t. That is full projectile motion, not the page that only computes Z.
A tower intuition check: v0 = 14 m/s at 30°, h0 = 12 m. Components are vx ≈ 12.1 m/s and vy ≈ 7.0 m/s. Height above ground is not the bare flat apex; it is h0 plus the rise vy²/(2g) ≈ 12 m + 2.5 m. Range comes only after time from y(t) = 0, not from Z = v0² sin(2α)/g.
Units: speed in m/s, height in meters, time in seconds, g in m/s². Angle α in problem text is usually in degrees. Convert km/h to m/s (divide by 3.6) before substituting. If you type 36 km/h as though it were already m/s, the result will be nonsense: that is 10 m/s, not thirty-six.Units: speed in ft/s, height in feet, time in seconds, g in ft/s². Angle α in problem text is usually in degrees. Convert mph to ft/s before substituting. If you type an mph value as though it were already ft/s, the result will be nonsense: 36 km/h is 10 m/s in SI, not “thirty-six feet per second”.
Common mistakes: (1) use Z = v0² sin(2α)/g when h0 ≠ 0, (2) mix H_max with Z, (3) take flight time from rise only instead of the full fall, (4) forget that a horizontal throw has no vertical launch component.
A sharp version of that mistake: someone takes v0 = 20 m/s at 45° from a balcony h0 = 15 m and computes Z = 400/9.81 ≈ 40.8 m with the flat formula. The path lands below the launch, so flight time is longer than 2·v0·sinα/g and the real range is clearly larger. The flat formula does not close a balcony problem.
Which calculator: vertical → vertical throw; roof or bridge with horizontal v → horizontal throw; mainly Z on flat ground → projectile range; mainly H_max → projectile height; h0, path, components → full projectile motion.
A vertical throw is the limit vx = 0. A horizontal throw is the limit vy0 = 0 with y0 = h > 0. An angled throw on flat ground is y0 = 0 and landing again at y = 0. The full path allows any h0. Naming the limit protects you from the wrong formula.
The model has limits. Air drag shortens the flight and breaks parabola symmetry, especially for a light ball and large v0. Wind adds a component the table does not know. Landing on a slope is not y = 0 on flat ground. Then the rows below are only a sketch or the wrong tool.
Before you compute, check in order: (1) whether v0 is vertical, horizontal or at an angle, (2) whether launch and landing share a height, (3) whether the problem uses g = 9.81 m/s² or 10 m/s², (4) whether units match the page header choice, (5) whether the result makes sense (at 10 m/s, heights and ranges of a few meters, not kilometers).
Details and worked examples for each model are in parts 4-6 of Physics without mysteries. This page is the comparison: launch conditions first, then the table row, then the calculator. For a map of the whole motion topic, see kinematics from scratch.
Read the problem: is v0 only vertical, only horizontal from a height, or at angle α? Check whether launch and landing are at the same height. Then open the matching table row and calculator.
When h0 > 0 and there is an angle α, do not use flat range. Use full projectile motion. For H_max in a pure vertical throw, vertical throw is enough.
Constant g, no air drag, independent horizontal and vertical motion. Launch: pure vertical, pure horizontal from a height, or v0 at angle α. We work in meters, seconds and m/s.Constant g, no air drag, independent horizontal and vertical motion. Launch: pure vertical, pure horizontal from a height, or v0 at angle α. We work in feet, seconds and ft/s.
When drag, wind or ball spin clearly change the path. When h0 > 0 and you still force Z = v0² sin(2α)/g. When you land on a slope instead of at y = 0. In those three cases the table is only a sketch or the wrong tool. Move to the full path or to the matching series article.
| Model | H_max | Z (range) | t (flight) | Key formula | When to use |
|---|---|---|---|---|---|
| Vertical throw | v₀²/(2g) | - | 2v₀/g (up and back) | H = v₀²/(2g) | Ball thrown straight up, fountain, jump |
| Horizontal throw | ≈ h₀ (start) | v₀·√(2h/g) | √(2h/g) | x = v₀ t, y = h − ½gt² | From edge, roof, bridge |
| Projectile (flat) | v₀² sin²α/(2g) | v₀² sin(2α)/g | 2v₀ sinα/g | Z = v₀² sin(2α)/g | Sport, α and v₀ problems |
| Full projectile | h₀ + vᵧ₀²/(2g) | vₓ·t (from y=0) | root of y=0 | h₀ + vᵧ₀ t − ½gt² = 0 | h₀ > 0, path, vₓ/vᵧ |
You throw a ball straight up at v0 = 10 m/s. Take g = 9.81 m/s². What is the maximum height H_max?
Steps
Answer: H_max ≈ 5.10 m
On flat ground, zero angle means the path “slides” on the ground and classic range as a flight loses meaning. A horizontal throw assumes launch from height h > 0 with only horizontal v. That is a different boundary condition than “zero angle on flat ground”.
When launch and landing are at the same height, g = const and there is no drag. From a tower or slope that formula does not close the problem. Then find time from y(t) = 0 with h0.
In the no-drag flat-ground model, Z grows with sin(2α). Sine peaks at 1 when 2α = 90°, so α = 45°. With drag the optimal angle is usually smaller.
In a typical horizontal throw there is no rise: vy at launch is zero, so maximum height above ground is the launch height h0 (then you only fall). H_max “above launch” appears when vy is positive.
First the rise: t_up = v0/g. Back to the same height, total flight is 2·t_up. If you land lower than the launch height, you need the full solution with a height offset, not bare 2·v0/g.
Use full projectile motion. The flat-ground range calculator will use the flat Z formula and ignore the tower. The full model takes h0, components and the path.
Yes if the problem says so or you agree that value for the whole calculation. Consistency matters. Tools often default to 9.81 m/s², so enter g = 10 m/s² deliberately when needed.
Underline the question first: “how high” or “how far horizontally”. H_max is height, Z is range. The table keeps separate columns. Then pick the model row and the matching calculator.
Yes, it usually shortens the flight and breaks parabola symmetry. This comparison does not compute that. If the problem says to neglect drag, stay with the table below.
In parts 4-6 of Physics without mysteries and next to the specific calculators. This page compares models; it is not a full problem set.