Projectile Motion Calculator
Launch speed, angle and height in; range, peak height, flight time and the full trajectory out, on Earth or any planet.
By Manzoor Ahmad, pharmacist · Updated How do you calculate projectile range and height?
From level ground: range R = v² sin 2θ ÷ g, maximum height H = v² sin²θ ÷ 2g and time of flight t = 2v sin θ ÷ g. A ball launched at 20 m/s and 45° on Earth flies 40.8 m, peaks at 10.2 m and stays in the air for 2.88 s, ignoring air resistance.
Projectile motion equations
| Quantity | From the ground (h₀ = 0) | From height h₀ |
|---|---|---|
| Horizontal velocity | vₓ = v cos θ (constant) | same |
| Vertical velocity | v_y = v sin θ − gt | same |
| Time of flight | t = 2v sin θ ÷ g | t = (v sin θ + √(v² sin²θ + 2gh₀)) ÷ g |
| Range | R = v² sin 2θ ÷ g | R = vₓ × t |
| Maximum height | H = v² sin²θ ÷ 2g | H = h₀ + v² sin²θ ÷ 2g |
| Best angle | 45° | θ = arctan(v ÷ √(v² + 2gh₀)) |
Range at 20 m/s by launch angle (Earth, ground level)
| Angle | Range | Max height | Flight time |
|---|
Notice the pairs: 15° and 75°, or 30° and 60°, land at the same spot. For a vertical-only problem use the free fall calculator.
Projectile motion FAQs
For launch speed v, angle θ and gravity g from ground level: time of flight t = 2v sin θ ÷ g, range R = v² sin 2θ ÷ g, and maximum height H = v² sin²θ ÷ 2g. Horizontal velocity v cos θ stays constant; vertical velocity changes at −g.
From level ground, R = v² sin(2θ) ÷ g. A ball launched at 20 m/s and 45° travels 20² × 1 ÷ 9.81 = 40.8 m, ignoring air resistance.
45° on level ground. When launching from a height h, the best angle is lower: θ = arctan(v ÷ √(v² + 2gh)). At 10 m/s from 10 m up it is about 30°.
H = h₀ + (v sin θ)² ÷ 2g. At 20 m/s and 45° from the ground, H = (14.14)² ÷ 19.61 = 10.2 m.
Solve h₀ + v sin θ · t − ½gt² = 0: t = (v sin θ + √((v sin θ)² + 2gh₀)) ÷ g. A ball rolled horizontally off a 10 m ledge lands after √(2 × 10 ÷ 9.81) = 1.43 s.
Because sin 2θ = sin(180° − 2θ), launches at 30° and 60° (or 20° and 70°) land at the same distance on level ground. The steeper one flies higher and stays up longer.
Not without air resistance. A bowling ball and a tennis ball launched with the same speed and angle follow the same path in a vacuum.
Drag shortens range and height, makes the descent steeper than the climb, and lowers the best angle below 45°, typically to 35–40° for balls and much lower for fast, light objects.
y = h₀ + x tan θ − g x² ÷ (2v² cos²θ). It is a parabola, which is why projectile paths are called parabolic.
Combine the constant horizontal speed with the final vertical speed: v_impact = √(vx² + vy²). From ground level it equals the launch speed; from a height it is faster.
About six times farther than on Earth for the same launch, since range is inversely proportional to g (1.62 vs 9.81 m/s²). Try the planet selector.
Yes. Treat it as two independent motions: constant velocity horizontally and constant acceleration (free fall) vertically, linked only by time.
Sources
Ideal projectile motion: no air resistance, spin or wind, and uniform gravity. Real balls and arrows fall short of these figures, especially at high speed.