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.
By AdminDrafted with AI writing toolsUpdated Checked examples: 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.
On level ground with no air resistance, the range is R = v² sin(2θ) ÷ g. A ball launched at 20 m/s and 45° on Earth (g = 9.80665 m/s²) lands 40.79 m away after 2.88 s, peaking at 10.20 m. From a raised launch point, find the flight time first instead.
Split the launch velocity into components and treat each direction separately, as in OpenStax University Physics, section 4.3. For the 20 m/s, 45° throw:
With no drag, the projectile lands at its launch speed, 20 m/s, and at 45° below the horizontal.
When the launch point is h above the landing ground, the flight time is the positive root of h + vy·t − ½gt² = 0: T = (vy + √(vy² + 2gh)) ÷ g. The range is still vx × T, and the peak is H = h + vy² ÷ 2g.
A stone thrown at 15 m/s and 30° from a 10 m cliff stays up 2.385 s, lands 30.98 m out and peaks 12.87 m above the base. It hits at 20.52 m/s, 50.73° below the horizontal.
A horizontal launch simply falls while it travels: T = √(2h ÷ g). At 10 m/s from 20 m, T = 2.020 s and the range is 20.20 m.
A throw angled downward lands sooner, and its highest point is the launch height. At 15 m/s and −20° from the same 10 m cliff, the stone lands 14.06 m out after 0.998 s.
On level ground, 45° throws farthest because sin 2θ peaks when 2θ = 90°. Complementary angles such as 30° and 60° land at the same spot, though the steeper one flies higher and longer.
A raised launch point moves the best angle below 45°, and so does air drag.
Range, height and flight time all scale with 1/g. The same 20 m/s, 45° throw with the surface gravities from NASA's planetary fact sheets:
Earth's figure is standard gravity. The local value varies slightly with latitude and altitude.
The linear drag model applies a force F = −b·v, with b in kg/s setting its strength. That b is a different quantity from the dimensionless drag coefficient used with v² drag.
The motion then depends on k = b/m. For a 0.145 kg ball with b = 0.01 kg/s, k = 0.069 s⁻¹ and the terminal speed g/k is 142.2 m/s.
Thrown at 20 m/s and 45°, that ball lands 35.95 m away instead of 40.79 m, about 12% shorter. It peaks at 9.57 m and comes down more steeply than it went up, at 48.7° instead of 45°.
Linear drag suits small, slow objects. A ball at sports speeds meets drag closer to v², so treat the drag result as a trend (Taylor, Classical Mechanics, sections 2.2 and 2.3).
Default is 20 m/s at 45° from ground level. Horizontal is 10 m/s at 0° from a 20 m height, and linear drag is the same 20 m/s, 45° throw with a 0.145 kg ball and b = 0.01 kg/s. All use Earth gravity.
| Scenario | Range (m) | Maximum height (m) | Flight time (s) |
|---|---|---|---|
| Default | 40.7886 | 10.1972 | 2.88419 |
| Horizontal | 20.1962 | 20 | 2.01962 |
| Linear drag | 35.9451 | 9.56778 | 2.79449 |
For straight-line motion with constant acceleration, use the kinematics calculator. When gravity weakens with distance, as for satellites, use the gravity and orbit calculator.
Checked against: R = v²sin2θ/g, H = v²sin²θ/(2g), T = 2v·sinθ/g with g = 9.80665
Checked against: T = (v_y + √(v_y² + 2gh))/g = 2.3847661…, R = v_x·T, v_impact = √(v² + 2gh)
Checked against: 400/1.62 and 2·20·sin45°/1.62 (NASA g_Moon = 1.62 m/s²)
Checked against: 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.
This calculator includes 6 worked examples with independently sourced answers. They run in the test suite, and you can run them here too.