What projectile motion is and why it splits into two problems
A projectile is anything moving under gravity alone after it has been launched — a thrown ball, a struck golf shot, a water jet, a shell, a stunt bike leaving a ramp. The defining feature is that nothing pushes it any more. Once it is airborne, the only force in this model is its weight, and weight points straight down.
That single fact is what makes the problem tractable: because gravity has no horizontal component, the horizontal motion has no acceleration at all. The projectile keeps whatever horizontal velocity it left with, for ever, until it lands. Meanwhile the vertical motion is exactly the constant-acceleration problem handled by the free fall calculator, just with a non-zero starting velocity. The two axes never talk to each other except through the shared clock.
So the recipe is always the same. Resolve the launch velocity into a horizontal part v₀cos θ and a vertical part v₀sin θ. Solve the vertical problem for the time of flight, because that is the only unknown the two axes share. Then multiply that time by the constant horizontal velocity to get the range. Everything on this page follows from those three sentences.
What the model leaves out is air. A real projectile also feels drag opposing its motion, and possibly lift from spin. Drag is not a small correction for fast or light objects: a table tennis ball struck hard travels a small fraction of the distance this model gives. Spin complicates it further — a driven golf ball carries backspin, and the lift that generates keeps it airborne long enough to fly further than the drag-free parabola, not shorter. Use the vacuum answer as an exact result for a vacuum and a teaching tool, not as ballistics.
The formula, including the case where you launch from a height
Take up as positive and put the origin at the launch point. Vertically, the height above the launch point after time t is v₀sin θ · t − ½gt². The projectile lands when that equals −y₀, the drop to the landing surface. Solving that quadratic and keeping the positive root gives the time of flight printed at the top of this page:
T = (v₀sin θ + √((v₀sin θ)² + 2gy₀)) / g
Set y₀ = 0 and the square root collapses to v₀sin θ, leaving T = 2v₀sin θ/g — the familiar result that the projectile spends as long coming down as going up. Multiply by the horizontal velocity and use the double-angle identity 2 sin θ cos θ = sin 2θ, and the level-ground range becomes R = v₀² sin 2θ / g.
That compact form carries two consequences worth memorising. First, sin 2θ peaks at θ = 45°, so on level ground 45° maximises range. Second, sin 2θ takes the same value at θ and at 90° − θ, so 30° and 60° give identical ranges — the 30° shot is a flat fast one and the 60° shot a high slow one, and they land in the same place.
Both of those results are properties of the level-ground case only. Launch from a height and the symmetry breaks: the projectile gets extra flight time on the way down that a flatter shot exploits better, so the optimum angle drops below 45°, and it drops further the greater the launch height is relative to v₀²/g. The angle sweep table the calculator generates uses your own inputs, so you can read the optimum off it rather than assume it.
Maximum height is simpler. The vertical velocity is zero at the apex, so from v² = u² + 2as the rise above the launch point is (v₀sin θ)²/(2g), and the height above the ground adds y₀. The impact speed is simpler still: by energy conservation it depends only on the launch speed and the drop, √(v₀² + 2gy₀), and not at all on the angle.
Worked example: a ball thrown at 25 m/s and 35° from a 2 m release height
Work it through by hand with g = 9.80665 m/s².
- Resolve the launch velocity. cos 35° = 0.81915 and sin 35° = 0.57358, so vx = 25 × 0.81915 = 20.479 m/s and vy = 25 × 0.57358 = 14.339 m/s.
- Find the vertical speed at impact. vy,impact = √(14.339² + 2 × 9.80665 × 2) = √(205.61 + 39.227) = √244.84 = 15.647 m/s.
- Time of flight. T = (14.339 + 15.647) ÷ 9.80665 = 29.986 ÷ 9.80665 = 3.0577 s.
- Range. R = vxT = 20.479 × 3.0577 = 62.62 m.
- Maximum height. H = 2 + 14.339² ÷ (2 × 9.80665) = 2 + 205.61 ÷ 19.6133 = 2 + 10.483 = 12.48 m, reached at t = 14.339 ÷ 9.80665 = 1.4622 s.
- Impact speed and angle. Speed = √(20.479² + 15.647²) = √(419.4 + 244.8) = √664.2 = 25.77 m/s, at arctan(15.647 ÷ 20.479) = 37.4° below the horizontal. Check it against energy: √(25² + 2 × 9.80665 × 2) = √(625 + 39.23) = √664.2 = 25.77 m/s. The two routes agree.
Notice the asymmetry that the launch height creates. The ball rises for 1.462 s and falls for 1.596 s, and it comes down steeper (37.4°) than it went up (35°). On level ground both halves would be equal and the impact angle would match the launch angle exactly.
How to read the results
Start with the time of flight, because everything else hangs off it. If it looks wrong, the vertical component is wrong, and that is almost always an angle entered in the wrong units or measured from the wrong reference. This calculator measures the angle from the horizontal, which is the physics convention; sights and mortars sometimes quote elevation from the vertical instead.
Compare the range against the angle sweep the calculator prints. If your angle is well away from the peak of that column, you are giving up distance for no benefit — and the sweep also tells you the second angle that reaches the same range, which on level ground is its complement. In sport that pair is a real choice: the flatter shot arrives sooner and steeper into a defence, the higher one hangs longer.
Read the impact angle when clearance or bounce matters. A shot that arrives at 15° will skid; one arriving at 60° will bite. And read the maximum height against whatever is overhead. The calculator reports the apex above the landing surface, including the launch height, which is the number that matters for clearing a wall or a ceiling.
Finally, apply the reality discount. In air, the drag force grows with the square of speed and always opposes motion, so for a projectile that generates no lift the real trajectory is shorter than this one, steeper on descent than on ascent, and asymmetric in a way the vacuum parabola never is. The faster and lighter the projectile, the larger the discount — and a spinning projectile that does generate lift, such as a driven golf ball, can instead exceed the vacuum range.
Range, height and flight time at 20 m/s from level ground
| Launch angle | Range (m) | Max height (m) | Flight time (s) |
|---|---|---|---|
| 15° | 20.39 | 1.37 | 1.056 |
| 30° | 35.32 | 5.10 | 2.039 |
| 40° | 40.17 | 8.43 | 2.622 |
| 45° | 40.79 | 10.20 | 2.884 |
| 50° | 40.17 | 11.97 | 3.125 |
| 60° | 35.32 | 15.30 | 3.532 |
| 75° | 20.39 | 19.03 | 3.940 |
Ranges pair up around 45°: 30° and 60° both give 35.32 m, 15° and 75° both give 20.39 m. Height and flight time do not pair — they rise monotonically with angle.
Assumptions this model makes, and what breaks when they fail
- No air resistance. The largest limitation by far. Real ranges for fast, light or spinning projectiles fall well short of these figures, and the descent is steeper than the ascent rather than symmetric.
- No lift or spin. A backspun golf ball or a curving football generates aerodynamic force perpendicular to its motion, which this model has no term for.
- Flat, level landing surface. The range reported is to the height you entered as the landing plane. On a slope, solve for where the parabola meets the slope line instead.
- Uniform gravity, non-rotating Earth. Fine for anything short of artillery. Long-range gunnery must also correct for Coriolis deflection and the change in g with altitude.
- Angle measured from the horizontal. Entering an elevation quoted from the vertical will silently give you the complement of the trajectory you wanted.
- Launch height is measured to the landing surface, not to the ground under the launcher. If the projectile lands on a roof or a green above the tee, the effective y₀ is smaller than the drop to your feet.
Related tools and where to go next
If your projectile has no horizontal component, the problem collapses to one dimension and the free fall calculator or the final velocity calculator will be quicker. If you need the energy the projectile carries — for impact assessment, or to compare a thrown object with a dropped one — feed the impact speed into the kinetic energy calculator, remembering that impact speed here depends only on launch speed and drop height, never on angle.
If you want the force that launched the projectile rather than the flight that followed, the F = ma calculator and the linear momentum calculator handle the launch impulse: a known change in momentum divided by the contact time gives the average force on the ball, the bat or the barrel.
For genuine external ballistics you need drag. That means a numerical integration using a drag coefficient, a projectile's ballistic coefficient and an air-density model — the standard reference tables (G1, G7 and their relatives) exist precisely because no closed-form solution survives once drag is included. Treat the parabola as the zeroth-order answer that tells you the right order of magnitude and the right qualitative behaviour, and reach for a ballistics solver when the answer has to be right to the metre.
Key terms
- Range
- The horizontal distance from launch point to landing point, measured along the ground. It is not the length of the curved path the projectile traces.
- Time of flight
- The interval between launch and landing. It is set entirely by the vertical motion and the launch height, and is unaffected by the horizontal velocity.
- Apex
- The highest point of the trajectory, where the vertical velocity is momentarily zero while the horizontal velocity continues unchanged.
- Complementary angles
- A pair of launch angles summing to 90°. On level ground and in a vacuum they produce identical ranges with different flight times and heights.