Enter launch speed, angle, and launch height to find time of flight, maximum height, and horizontal range — assuming constant gravity and no air resistance, g = 9.81 m/s².
Projectile motion is the classic demonstration that horizontal and vertical motion can be analyzed completely independently: gravity only ever acts vertically, so the horizontal velocity stays constant for the whole flight while the vertical velocity changes at a constant rate. This calculator splits the launch velocity into its two components and applies basic kinematics to each.
A launch speed v₀ at angle θ above the horizontal splits into a horizontal component vₓ = v₀cos(θ) and a vertical component v_y = v₀sin(θ). Because there is no horizontal force (ignoring air resistance), vₓ never changes during the flight — it is the same value at launch, at the apex, and at landing.
The vertical position follows y(t) = h₀ + v_y·t − ½g·t². Setting y(t) = 0 and solving the resulting quadratic for t (using the quadratic formula, keeping the positive root) gives the time of flight: t = [v_y + √(v_y² + 2g·h₀)] / g. When the launch height h₀ is zero, this simplifies to the familiar t = 2v_y/g.
The projectile reaches its apex when the vertical velocity momentarily equals zero, at t = v_y/g. Substituting that time into the position equation gives the maximum height, h₀ + v_y²/(2g). The range — how far it travels horizontally before landing — is simply the constant horizontal velocity multiplied by the total time of flight: R = vₓ·t.
Take v₀ = 30 m/s at θ = 45° from ground level (h₀ = 0). vₓ = v_y = 30·cos(45°) = 30·sin(45°) ≈ 21.21 m/s. Time of flight t = 2(21.21)/9.81 ≈ 4.33 s. Maximum height = 21.21²/(2·9.81) ≈ 22.94 m. Range = 21.21 × 4.33 ≈ 91.8 m — and for a flat launch, 45° always maximizes range for a given speed.
For a launch and landing at the same height, range R = v₀²sin(2θ)/g. Since sin(2θ) is maximized when 2θ = 90° (θ = 45°), that angle gives the longest range for a fixed launch speed. Launching from a height above the landing point shifts the optimal angle slightly below 45°.
No — this calculator uses the idealized kinematics model with no air drag, which is the standard assumption for introductory physics and FE exam problems. Real projectiles (especially light or fast-moving ones) experience drag that shortens both range and time of flight relative to this ideal calculation.
A negative h₀ represents landing below the launch point (for example, throwing something off a cliff toward a lower target). The time-of-flight formula still applies as written and correctly returns a longer flight time than a level launch at the same speed and angle.
Finding the time of flight means solving −½g·t² + v_y·t + h₀ = 0 for t, which is a quadratic equation in t. This calculator's time-of-flight formula is exactly the quadratic formula applied with a = −g/2, b = v_y, and c = h₀ — see the Quadratic Equation & Root Finder for the general form.
Try our STEM Learning Studio
More calculators, simulators, and guides for this discipline.
Visual Engineering System Builder — STEM Fundamentals
✨ Premium ContentA structured, paid professional training program for this discipline.