Projectile Range (Flat Ground)

Projectile range: R = v₀²·sin(2θ) / g

Launch a projectile across a horizontal plane with no air resistance and the horizontal range comes out to R = v₀²·sin(2θ) / g, taking g ≈ 9.81 m/s². Since sin(2θ) hits its maximum at 2θ = 90°, the range is longest at θ = 45°. Try v₀ = 20 m/s and θ = 45° and you get R ≈ 40.8 m. Fire at 30° or 60° and the range matches at about 35.3 m, because sin(60°) = sin(120°). Bring in real ballistics with air drag and that optimal angle slides down to roughly 35°, with the trajectory turning asymmetric as the descent steepens. Fast projectiles such as rifle bullets fly flatter still, since drag dominates.

Applications

Ballistics for small arms and artillery, sports like free kicks, javelin, shot put and golf, rocket trajectory work in aerospace, mortar firing tables in defense, and even Formula 1 aerodynamics, where rear-wing downforce trades off against drag and shapes top speed.

FAQ

Why is 45° optimal only in vacuum? The formula leaves drag out entirely. Once air resistance is in play, steeper angles burn more energy fighting gravity while the horizontal speed stays low, so the sweet spot moves down to around 35°.

Do θ and 90° − θ give the same range? Yes. Without drag, 30° and 60° land at the same distance, because sin(2·30°) = sin(2·60°) = sin(60°).

How do I find range from a height (not flat ground)? Reach for the general formula R = (v₀·cos θ / g)·(v₀·sin θ + √((v₀·sin θ)² + 2gh)). Set h = 0 and it collapses back to the simple version.