Projectile motion calculator: range, maximum height, time of flight and impact speed for any launch angle and height, on Earth or other worlds, with drag.
Atualizado Exemplos verificados: 6
The projectile flies for 2.884 s, peaks 10.2 m above the ground and lands 40.79 m away at 20 m/s.
The launch velocity splits into a horizontal part, v cos θ, which stays constant, and a vertical part, v sin θ, which gravity reduces by g every second. The time of flight is when the height returns to the ground, found by solving the vertical motion; the range is horizontal speed × flight time. On level ground this reduces to R = v² sin 2θ ÷ g.
It answers homework problems and sports or water-jet estimates. The default throw, 20 m/s at 45° on Earth, peaks at 10.20 m and lands 40.79 m away after 2.88 s. The same throw on the Moon, where g = 1.62 m/s², travels 246.9 m.
Without the drag option, air resistance, spin and the Earth's curvature are ignored, which overstates the range of fast or light objects. The linear-drag option shows how drag shortens and steepens the path; real balls at sports speeds meet drag closer to v², so read that result as a trend.
Fonte de verificação: Python 3.8 math: R = v²sin2θ/g, H = v²sin²θ/(2g), T = 2v·sinθ/g with g = 9.80665
Fonte de verificação: Python 3.8 math: T = (v_y + √(v_y² + 2gh))/g = 2.3847661…, R = v_x·T, v_impact = √(v² + 2gh)
Fonte de verificação: Python 3.8 math: 400/1.62 and 2·20·sin45°/1.62 (NASA g_Moon = 1.62 m/s²)
Fonte de verificação: Python 3.8 math: T = √(2h/g) = 2.0196200…, R = vT, tan φ = gT/v
On level ground with no air resistance, 45° gives the longest range, because R = v² sin 2θ ÷ g and sin 2θ peaks when 2θ = 90°. Launching from a height moves the best angle lower: from a 10 m cliff at 15 m/s it is about 36.2°, which reaches 31.4 m against 30.5 m at 45°. Air drag also lowers the best angle.
On level ground, T = 2v sin θ ÷ g, so a 20 m/s launch at 45° on Earth stays up 2.88 s. From a launch height h, take the positive root of h + v sin θ·t − ½gt² = 0, which is T = (v sin θ + √(v² sin² θ + 2gh)) ÷ g. The time to the highest point is v sin θ ÷ g, half the level-ground flight time.
Complementary angles give the same range on level ground because sin 2θ = sin(180° − 2θ). At 20 m/s, launches at 30° and 60° both land 35.32 m away. The 60° shot climbs three times as high, 15.30 m against 5.10 m, and stays up 3.53 s instead of 2.04 s, which matters when the path has to clear an obstacle.
It depends on the object's mass, size and speed. With the linear model here, a 0.145 kg ball thrown at 20 m/s and 45° with b = 0.01 kg/s lands 35.95 m away instead of 40.79 m, about 12% shorter, and peaks at 9.57 m instead of 10.20 m. Drag also makes the descent steeper than the climb, so the path is no longer a symmetric parabola.
A precisão depende dos dados inseridos e das hipóteses do método. O cálculo decimal usa 50 algarismos significativos, mas estimativas, métodos numéricos e dados de origem podem ter menor precisão; o arredondamento exibido não elimina essas limitações. Exemplos resolvidos verificados com fontes independentes: 6. Por exemplo, “20 m/s at 45° on Earth” é verificado com Python 3.8 math: R = v²sin2θ/g, H = v²sin²θ/(2g), T = 2v·sinθ/g with g = 9.80665.
OpenStax University Physics Volume 1, §4.3 Projectile motion; Taylor, Classical Mechanics (2005), §2.2–2.3 Linear air resistance; NASA Planetary Fact Sheet — surface gravity.
Esta calculadora inclui 6 exemplos resolvidos com respostas de fontes independentes. Eles fazem parte do conjunto de testes e você também pode executá-los aqui.
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