Free Projectile Motion Calculator
Calculate the range, maximum height, and time of flight of a projectile launched from flat ground. Enter the initial speed and launch angle and get instant, formula-transparent results using g = 9.80665 m/s².
Enter launch speed and angle to find range, max height, and flight time (flat ground, g = 9.80665 m/s²).
R = v₀²·sin(2θ)/g · H = (v₀·sinθ)²/(2g) · T = 2·v₀·sinθ/g. Results assume launch and landing at the same height and ignore air resistance.
Quick answer
For a projectile launched and landing at the same height, range is R = v₀²·sin(2θ)/g, maximum height is H = (v₀·sinθ)²/(2g), and time of flight is T = 2·v₀·sinθ/g, with g = 9.80665 m/s². Launching at v₀ = 20 m/s and θ = 45° gives R ≈ 40.79 m, H ≈ 10.20 m, and T ≈ 2.884 s. The range is greatest at 45° because sin(2θ) peaks there.
Formula & method
Range (horizontal distance)
R = v₀² · sin(2θ) / g
- R — Range — horizontal distance (m)
- v₀ — Initial launch speed (m/s)
- θ — Launch angle above horizontal (degrees)
- g — Gravitational acceleration, 9.80665 m/s²
Horizontal distance traveled before landing at the same height. Maximized at θ = 45°, where sin(2θ) = 1.
Maximum height
H = (v₀ · sinθ)² / (2g)
- H — Maximum height above launch point (m)
Peak height reached, set by the vertical velocity component v₀·sinθ only.
Time of flight
T = 2 · v₀ · sinθ / g
- T — Total time of flight (s)
Total time in the air for launch and landing at equal height; it is twice the time to reach the apex.
Examples
- Input
- v₀ = 20 m/s, θ = 45°
- Result
- R ≈ 40.79 m, H ≈ 10.20 m, T ≈ 2.884 s
- Why
- sin(2·45°) = sin90° = 1, so R = 20²·1/9.80665 = 400/9.80665 ≈ 40.79 m. The vertical component is v₀·sin45° = 20·0.7071 = 14.142 m/s, so H = 14.142²/(2·9.80665) = 200/19.6133 ≈ 10.20 m and T = 2·14.142/9.80665 ≈ 2.884 s.
- Input
- v₀ = 30 m/s, θ = 30°
- Result
- R ≈ 79.48 m, H ≈ 11.47 m, T ≈ 3.059 s
- Why
- sin(2·30°) = sin60° = 0.8660, so R = 30²·0.8660/9.80665 = 779.4/9.80665 ≈ 79.48 m. Vertical component v₀·sin30° = 30·0.5 = 15 m/s gives H = 15²/(2·9.80665) = 225/19.6133 ≈ 11.47 m and T = 2·15/9.80665 ≈ 3.059 s.
- Input
- v₀ = 15 m/s, θ = 60°
- Result
- R ≈ 19.87 m, H ≈ 8.60 m, T ≈ 2.649 s
- Why
- sin(2·60°) = sin120° = 0.8660, so R = 15²·0.8660/9.80665 = 194.85/9.80665 ≈ 19.87 m. The vertical component v₀·sin60° = 15·0.8660 = 12.99 m/s yields H = 12.99²/(2·9.80665) ≈ 8.60 m and T = 2·12.99/9.80665 ≈ 2.649 s — note 30° and 60° share the same range but the 60° shot flies higher and hangs longer.
When to use this tool
- Physics homework and exam prep covering kinematics, two-dimensional motion, and projectile trajectories.
- Quickly checking range, apex height, or hang time for a ball, water jet, or launched object on level ground.
- Exploring how launch angle changes the trade-off between range and height — for example, why 45° maximizes distance.
- Sanity-checking a simulation or a more detailed model before adding air resistance or height offsets.
Common mistakes
- Entering the angle in radians instead of degrees. The calculator expects θ in degrees (e.g. 45), not 0.785 radians.
- Using sin(θ) where sin(2θ) belongs. Range uses sin(2θ); only the height and time formulas use sin(θ). Mixing them up is the most common arithmetic error.
- Assuming maximum height occurs at 45°. The 45° angle maximizes range, but maximum height keeps increasing toward 90° (a straight-up shot), where range is zero.
- Applying these formulas when launch and landing heights differ (e.g. a cliff or a basketball shot). The R, H, T equations here assume the projectile returns to its starting height; uneven heights need the full quadratic for vertical motion.
Frequently asked questions
What launch angle gives the maximum range?
On flat ground with no air resistance, 45° gives the maximum range because R = v₀²·sin(2θ)/g and sin(2θ) reaches its peak value of 1 when 2θ = 90°, i.e. θ = 45°. Complementary angles such as 30° and 60° produce the same range, but the steeper one flies higher and stays airborne longer.
Does this calculator account for air resistance?
No. It uses the idealized projectile equations that ignore drag, wind, and spin. Real-world ranges are usually shorter than the calculated value, especially for light objects or high speeds. For most classroom problems and quick estimates, the no-drag model is the expected answer.
What value of gravity does it use?
It uses g = 9.80665 m/s², the standard gravity defined by the International Bureau of Weights and Measures. Local gravity varies slightly with latitude and altitude (roughly 9.78–9.83 m/s²), but 9.80665 is the conventional value used in physics calculations.
How do I convert km/h to m/s for the speed input?
Divide the speed in kilometres per hour by 3.6. For example, 72 km/h ÷ 3.6 = 20 m/s. The calculator works entirely in SI units, so speeds must be in metres per second and the result distances and heights come out in metres.
Why are the time of flight and max height the same for 30° and 60°? (They aren't.)
They aren't — only the range matches for complementary angles. A 30° and a 60° launch at the same speed share an identical range, but the 60° shot has a larger vertical velocity component (v₀·sinθ), so it reaches a higher peak and a longer time of flight. The 30° shot is flatter and faster horizontally.
Can I use it for a projectile launched from a height?
Not directly. These formulas assume the projectile lands at the same height it was launched from. If you fire from a cliff or a table, the actual flight time and range are longer, and you need the full kinematic equation y = v₀·sinθ·t − ½gt² solved for the real landing height. Use this tool for level-ground problems.
Sources & references
- HyperPhysics — Trajectory (Projectile Motion)
- Wikipedia — Projectile motion
- NIST — Standard acceleration of gravity (gₙ = 9.80665 m/s²)
External references open in a new tab. We are independent and not affiliated with these organizations.
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