Projectile Motion Calculator
Range, height and flight time from any launch — with air drag and a target solver.
Trajectory
Position over time
How it was worked out
About the Projectile Motion Calculator
Enter a launch speed, angle and height and get the time of flight, the range, the maximum height (and when and where it is reached), and the speed and angle at impact — on Earth, the Moon, another planet or with your own g. The trajectory is plotted, a table gives the position over time, and the working shows every formula with your numbers.
Turn on air resistance to add quadratic drag (½ ρ Cd A v²) from the object’s mass, drag coefficient, size and the air density. The motion is then integrated numerically with a fourth-order Runge–Kutta method and compared with the vacuum path, together with the object’s terminal velocity.
Launch angle to hit a target works out the angles — usually a low and a high one — that reach a given distance (and height), or tells you the smallest speed that could get there.
How to use it
- Type the launch speed and pick its unit, then the angle above the horizontal (negative to aim down) and the launch height above the ground where it lands.
- Pick the gravity. Earth’s standard g is 9.80665 m/s²; textbook problems often use 9.81 or 9.8.
- Read the range, time of flight, maximum height and impact speed, and look at the trajectory. Tick True proportions to see the real shape of the path.
- Open Air resistance and tick the box to add drag: enter the mass, the drag coefficient (or pick a typical one), the diameter or area, and the air density.
- Open Launch angle to hit a target, type the distance (and height), and press Use this angle next to the angle you want.
Examples
20 m/s at 45°, from the ground
Range 40.79 m (= v² ÷ g), time of flight 2.884 s, maximum height 10.20 m
15 m/s horizontally from 100 m (g = 9.8)
Lands 67.76 m out after 4.518 s, at 46.74 m/s, 71.3° below the horizontal
30 m/s at 45°, landing 10 m above the launch point (height −10 m, g = 9.8)
Time of flight 3.79 s; impact at 26.5 m/s, 36.9° below the horizontal
20 m/s, target 30 m away on level ground
23.67° (low) or 66.33° (high); it needs at least 17.15 m/s
150 g ball, 7 cm across, Cd 0.45, thrown at 30 m/s at 45°
Range 62.53 m instead of 91.77 m in a vacuum; impact at 21.39 m/s, 54.3° below the horizontal
the same ball at 30 m/s
about 42° gives the longest range (62.8 m) — less than the 45° of a vacuum
Common uses
- Physics homework on projectiles launched from the ground, from a height, or onto higher ground — with the working.
- Sport: how far a throw, kick or hit goes, and how much air resistance shortens it.
- Finding the angle needed to reach a target at a known distance and height.
- Comparing gravity on the Moon or Mars with Earth’s.
Formulas without air resistance
Split the launch velocity v₀ at angle θ into v₀x = v₀ cos θ and v₀y = v₀ sin θ. Horizontally the speed stays constant; vertically the projectile accelerates downwards at g. With a launch height h₀ above the landing level:
- Time of flight:
T = (v₀y + √(v₀y² + 2 g h₀)) ÷ g— on level groundT = 2 v₀ sin θ ÷ g - Range:
R = v₀x × T— on level groundR = v₀² sin 2θ ÷ g, largest at 45°, and the same for angles that add up to 90° - Maximum height:
H = h₀ + v₀y² ÷ 2g, reached att = v₀y ÷ g - Impact:
vy = v₀y − g T, speed√(v₀x² + vy²) - Angle to a target x away and y higher than the launch:
tan θ = (v² ± √(v⁴ − g(g x² + 2 y v²))) ÷ (g x). If the square root is of a negative number, the target is out of reach.
How air drag is modelled
The drag force is F_D = ½ ρ C_d A v², pointing against the velocity (NASA Glenn’s drag equation). Dividing by the mass gives the equations of motion dvx/dt = −k v vx and dvy/dt = −g − k v vy with k = ρ C_d A ÷ 2m and v = √(vx² + vy²). They have no closed-form solution, so the calculator integrates them with fourth-order Runge–Kutta in small time steps and pins down the landing and the top of the arc inside the final step.
With drag, the descent is steeper than the climb, the impact is slower than the launch, and the best angle for distance is below 45°. Typical drag coefficients (OpenStax Table 6.2) are 0.45 for a sphere, 0.70 for a skydiver feet first and 1.0 spread-eagle; NASA Glenn gives 0.75 for a model rocket. Real balls also spin and their Cd changes with speed, so treat drag results as estimates.
Sources
- OpenStax, University Physics Volume 1: §4.3 Projectile Motion (time of flight Eq. 4.24, range Eq. 4.26, maximum height, Example 4.8) and §6.4 Drag Force and Terminal Speed (Eq. 6.5 and Table 6.2).
- NASA Glenn Research Center: The Drag Equation, Terminal Velocity and Shape Effects on Drag.
- Air density 1.225 kg/m³ at sea level and 15 °C: the U.S. Standard Atmosphere, 1976 (NASA-TM-X-74335); NASA JPL Planetary Physical Parameters for g on other bodies.
Limitations
- Flat ground and constant g: fine for sport and most homework, but not for ranges of many kilometres, where Earth’s curvature and the change of g with height matter.
- Drag uses one constant drag coefficient. It ignores spin (the Magnus effect behind curve balls), wind, and the large rise in Cd near the speed of sound.
- The air density preset is Earth’s; set it for other atmospheres, or 0 for the Moon.
- With drag, target angles are found numerically: a 2° scan finds them and bisection refines them, with an extra search around the best angle for targets close to the longest reach. They are accurate for the model, so only as good as the drag values you enter.
Privacy
Everything happens in your browser. What you enter or open here is not uploaded or stored by MySmartCoPilot.
Frequently asked questions
How do you calculate the range of a projectile?
On level ground without air resistance, R = v₀² sin 2θ ÷ g. At 20 m/s and 45°, R = 400 × 1 ÷ 9.807 = 40.79 m. From a height, first find the time of flight T = (v₀y + √(v₀y² + 2gh₀)) ÷ g, then R = v₀ cos θ × T.
What angle gives the maximum range?
45° on level ground without air resistance, and angles that add up to 90° (say 30° and 60°) give the same range. Launching from a height lowers the best angle, and so does air drag — for the default 150 g ball at 30 m/s it is about 42°.
How do I find the maximum height?
At the top the vertical velocity is zero, so H = h₀ + (v₀ sin θ)² ÷ 2g, reached after v₀ sin θ ÷ g seconds. At 20 m/s and 45° from the ground: (14.14)² ÷ 19.61 = 10.20 m after 1.442 s.
Does the mass of the projectile matter?
Not without air resistance: every object falls with the same acceleration g. With air drag it does — a heavier object of the same size and shape has a smaller k = ρ Cd A ÷ 2m, so it slows down less and goes farther.
Why are there two launch angles for one target?
Because a lower, faster path and a higher, slower lob can both pass through the same point. At 20 m/s, a target 30 m away is reached at 23.67° or 66.33°. If the target is beyond the maximum range, neither exists.
How accurate is the air-resistance mode?
The numerical integration is accurate to well under 0.1% for the model; the uncertainty is in the inputs. The drag coefficient of a real object depends on its exact shape, surface and speed, and spin and wind are not included, so treat the drag results as an estimate.