Motion · Mechanics

Projectile Motion Calculator

Calculate initial velocity components, flight time, horizontal range, maximum height, and the ideal trajectory from ground level.

Formula shown Runs locally Reviewed Aug 11, 2026

Equation guide

Ideal launch from ground level; constant g; no air resistance

Initial velocity components

v₀ₓ = v₀ cos θv₀ᵧ = v₀ sin θ

Laws of motion

x(t) = v₀ cos θ · ty(t) = v₀ sin θ · t − gt²/2

Laws of velocity

vₓ(t) = v₀ cos θvᵧ(t) = v₀ sin θ − gt

Trajectory equation

y(x) = x tan θ − gx²/(2v₀² cos² θ)

Safety parabola

yₛ(x) = v₀²/(2g) − gx²/(2v₀²)

Flight characteristics

T = 2v₀ sin θ/gR = v₀² sin(2θ)/gH = v₀² sin² θ/(2g)

Mechanics calculator

Ideal projectile from ground level

Initial speed (m/s)

Manual entry: up to 3 decimal places; scientific notation is accepted with the same mantissa rule · button step: 0.5 m/sLaunch angle from horizontal (°)

Manual entry: up to 3 decimal places; scientific notation is accepted with the same mantissa rule · button step: 0.5 °

v₀ · θ

Velocity components, flight time, range, height, and safety-parabola limits will appear here.

Trajectory graph

Enter the initial speed and launch angle, then calculate to draw the ideal path.

Orthogonal coordinate systemThe calculated trajectory will appear here.

Using the model

Symbols, assumptions and limitations

Interpretation notes

The selected trajectory y(x) follows by eliminating time from x(t) and y(t). The safety parabola yₛ(x) = v₀²/(2g) − gx²/(2v₀²) is the envelope of every ideal trajectory launched from the origin with the same initial speed but any launch angle: points above it cannot be reached at that speed. Both models assume constant g = 9.80665 m/s² and negligible air resistance.

Worked examples

See the method in practice

Launch at 45°

At 10 m/s and 45°, v₀x = v₀y ≈ 7.071 m/s and the ideal horizontal range is about 10.20 m.

Safety boundary

For v₀ = 10 m/s, the safety parabola has its highest point at v₀²/(2g) ≈ 5.10 m and meets Ox at the maximum possible range v₀²/g ≈ 10.20 m.

Questions

Frequently asked

What does the trajectory equation describe?

It gives vertical position y directly as a function of horizontal position x. Under the stated assumptions, its graph is a downward-opening parabola.

What is the safety parabola?

It is the envelope of all ideal trajectories with the same launch point and initial speed as the angle varies. A point above this boundary cannot be reached with that initial speed; a point below it can generally be reached using one or two launch angles.

Why is air resistance omitted?

A useful drag model requires shape, area, drag coefficient, air density, and numerical integration.

Is 45° always optimal?

Only for equal launch and landing heights without drag.

Sources & review

Equations and examples are checked against the references below. Results are educational and should be independently verified for safety-critical work.

OpenStax Physics — projectile motion

Written by the STEM Hub editorial team · Reviewed August 11, 2026 · Review methodology