Projectile Launched From a Height Calculator
Launch a projectile from an elevated point and find its range, flight time, peak height and impact speed for any launch angle.
How it works
vₓ = v₀·cosθ · v_y = v₀·sinθ · t = (v_y + √(v_y² + 2gh)) ÷ g · range = vₓ·t · impact speed = √(v₀² + 2gh)
A projectile fired from a height keeps a constant horizontal velocity while gravity acts only vertically. The extra launch height lengthens the fall, so the object stays airborne longer than a ground-level shot and lands farther away. Impact speed follows from energy conservation — √(v₀² + 2gh) — and is independent of launch angle, because the extra drop converts the same potential energy into kinetic energy regardless of direction.
Worked example
At 20 m/s, 30° above horizontal, from 10 m up with g = 9.81: vₓ = 17.32 m/s, v_y = 10 m/s. Flight time t = (10 + √(100 + 196.2)) ÷ 9.81 ≈ 2.77 s, so range = 17.32 × 2.77 ≈ 48.0 m. Impact speed = √(400 + 196.2) ≈ 24.4 m/s.
Sources: Projectile motion — Wikipedia
Frequently asked questions
Why does the projectile go farther than one launched from the ground?
The launch height adds fall distance after the trajectory returns to the starting level, keeping the projectile in the air longer. Because horizontal velocity is constant, more airtime means more horizontal distance.
Is 45° still the best angle when launching from a height?
No. From an elevation the optimum angle for maximum range is less than 45°, and it decreases as the launch height grows relative to the speed. The extra height rewards a flatter, faster horizontal component.
Does the impact speed depend on the launch angle?
No. Impact speed is √(v₀² + 2gh), which contains no angle. Two projectiles launched at the same speed from the same height land at the same speed, though at different points and angles.
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This calculator is for educational purposes. Double-check important results before acting on them.