Projectile Range Calculator with Launch Height
Written by Thierno Sadou Diallo, formula verified per our methodology • Last checked on 9/3/2026
For a projectile launched from a height h0, the range is found by first solving for the flight time t = [v0×sin(θ) + √((v0×sin(θ))² + 2×g×h0)] ÷ g, then R = v0×cos(θ)×t. For a speed of 20 m/s at 30° from a height of 10 m, the range reaches about 48.04 m — noticeably farther than the 35.31 m obtained with no launch height, since the extra falling time lets the projectile cover a greater horizontal distance.
Explanation
The classic parabolic-trajectory formula, R=v²sin(2θ)/g, only applies to the special case where the projectile lands at exactly the same level it was launched from. As soon as the launch point is higher than the landing point — an extremely common scenario in practice (a ball thrown from a balcony, a projectile fired from a hill, a golf ball hit from an elevated tee) — this simple formula is no longer enough, and a genuine second-degree equation must be solved to find the exact flight time. This calculator carries out that full calculation in several steps: it first breaks the initial velocity down into its horizontal and vertical components, then solves for the total flight time accounting for the extra falling height, and finally derives the horizontal range and the maximum height reached by the projectile. A simple, reassuring consistency check on this more general formula: setting the initial height to zero gives back EXACTLY the classic parabolic-trajectory formula, and the underlying physics connects directly to concepts covered by our kinetic energy calculator (the projectile trades height for speed throughout its flight) and by Newton's second law (the constant downward acceleration g driving the whole motion) — which is also why a positive launch height always makes the range longer than the no-height case, at the same speed and angle: the projectile simply has more time to cover horizontal distance before hitting the ground.
Example: a ball thrown from a balcony
Inputs
Initial speed v0 = 20 m/s. Launch angle θ = 30°. Initial height h0 = 10 m (an upper-floor balcony).
Calculation
vx=20×cos(30°)≈17.32 m/s. vy=20×sin(30°)=10 m/s. t=(10+√(100+2×9.81×10))÷9.81≈2.774 s. R=17.32×2.774≈48.04 m.
Result
This ball travels about 48.04 m horizontally before hitting the ground, and reaches a maximum height of about 15.10 m above the ground during its flight — a range noticeably greater than the 35.31 m it would have covered launched from ground level, at the same speed and angle.
Frequently asked questions
Why does the flight time formula involve a square root?
This square root comes directly from solving the second-degree equation describing the projectile's height over time (h0+vy×t−½gt²=0): the standard quadratic formula (−b±√(b²−4ac))÷(2a) naturally produces a square root. Only the root corresponding to a positive time makes physical sense (the other root would correspond to a negative time, before launch, which has no real existence for this problem) — which is why the formula only keeps the + sign in front of the root.
Why does a higher launch height always increase the range, at the same speed and angle?
A higher initial height means the projectile has farther to fall before hitting the ground, which mechanically extends its total flight time. Since the horizontal range is the product of the horizontal speed (constant throughout the flight, absent air resistance) and this flight time, a longer flight time directly translates into a greater range, regardless of the speed or angle chosen.
Does this formula account for air resistance?
No, like nearly all classic ballistics formulas taught in general physics, this calculator assumes motion in a vacuum, with no air resistance at all. For a light or fast projectile (a hard-hit golf ball, for example), air resistance noticeably reduces the real range compared to this theoretical estimate, an effect whose size depends heavily on the shape, mass, and speed of the projectile in question.