How this instrument works
The trajectory equation answers a more specific question than most projectile calculators attempt: not how far something travels or how high it climbs, but where it actually is, its height y, at any horizontal distance x you name. It comes from combining the two independent motions of a launched object: horizontal position grows steadily as x = v0cosθ·t, while vertical position follows y = v0sinθ·t − ½gt². Eliminating time between them, by substituting t = x ⁄ (v0cosθ), turns sinθ⁄cosθ into tanθ and leaves g in a term divided by cos²θ, which is exactly the shape you see in the formula box.
The first term, x·tanθ, is where the projectile would be if gravity switched off, a straight line at the launch angle. The second term, gx² ⁄ (2v0²cos²θ), is how far gravity has pulled the projectile down from that straight line by the time it has covered distance x, since a longer horizontal trip means more time falling. Ballisticians call this the vacuum or ideal trajectory; an artillery officer, an archer sighting past 40 m, and a mechanical-engineering student checking a nozzle's spray pattern all lean on the same parabola, just with different names for v0 and θ.
The formula assumes level ground and no air resistance, so it is exact for that idealized case and only approximate for a real bullet, arrow, or thrown ball once drag starts to matter, bending the true path below this parabola more sharply as speed rises. It also has no opinion about where the ground actually sits: a negative y at some x is a perfectly valid answer, meaning that point on the curve is below the launch height, which is exactly what you want when firing from a cliff edge or down into a valley.
- Enter Launch speed — the projectile's velocity the instant it leaves the barrel, bow, or hand, in m/s.
- Enter Launch angle — degrees measured up from horizontal; 45° gives the longest range once air drag is ignored.
- Enter Horizontal distance — how far downrange, in metres, you want to check the projectile's height.
- Read Height at that distance — positive means above the launch point, negative means below it.
- Change Horizontal distance and re-check Height at that distance to trace the whole arc, from launch to landing.
Worked example — height 50 m downrange from a 45° launch
Launch speed v0 = 30 m/s, Launch angle θ = 45°, Horizontal distance x = 50 m. At 45°, tanθ = 1 exactly, so the straight-line term x·tanθ is just 50 m. Also at 45°, cos²θ = 0.5, so the denominator 2v0²cos²θ becomes 2 × 900 × 0.5 = 900. The gravity term's numerator is gx² = 9.80665 × 2500 = 24516.625, and dividing that by 900 gives 27.240694 m of drop.
Subtracting the drop from the straight-line term, y = 50 − 27.240694 = 22.759306 m, so the instrument's Height at that distance reads 22.7593 m. That is roughly the height of a seven-storey building, reached 50 m out from a shot that, by the standard range formula v0²sin(2θ)⁄g, lands again around 91.8 m downrange, and this trajectory equation is what tells you the projectile's altitude anywhere in between, not just at that landing point.
Questions
How is this different from a calculator that only gives range or max height?
It gives height at any horizontal distance you choose, not just the two special points other tools report. A range calculator only tells you where the parabola crosses zero again, and a max-height calculator only finds the peak; this instrument evaluates y = x·tanθ − gx² ⁄ (2v0²cos²θ) at whatever x you enter, so you can check the shot's altitude at 10 m, 50 m, or 90 m along the same flight.
Where does the tanθ and cos²θ in the formula actually come from?
From eliminating time between the two independent equations of motion. Horizontal position is x = v0cosθ·t and vertical position is y = v0sinθ·t − ½gt²; solving the first for t = x ⁄ (v0cosθ) and substituting into the second turns sinθ⁄cosθ into tanθ and squares the cosθ underneath the gravity term. Nothing is added or approximated in that step, it is the same parabola, just written as a function of distance instead of time.
Does the formula account for air resistance?
No, it describes the ideal vacuum trajectory, a pure parabola. Real bullets, arrows, and thrown balls fall faster than this equation predicts once drag becomes significant, curving below the plotted path, and the gap grows with speed and with how blunt the object is. For low-speed, short-range motion, a tossed ball, a thrown wrench, an arrow at target-archery ranges, the vacuum trajectory is close enough that ballistics texts still teach it first.
What does a negative Height at that distance mean?
That the point you asked about sits below the launch height, not that anything went wrong. It happens naturally past the point where the projectile returns to launch level, or immediately if you are firing from a height down into a valley or off a cliff, since the equation keeps describing the same parabola no matter where the actual ground happens to be.
Why does 45° matter so much for Launch angle?
Because on level ground with no air resistance, 45° maximizes total range for a given Launch speed, splitting the launch velocity evenly between horizontal and vertical components. Angles above or below 45° by the same amount, like 30° and 60°, land at the same range as each other but trace different arcs, so this instrument lets you compare their heights partway along the flight rather than only their endpoints.
Is this accurate enough for real artillery or long-range shooting?
For short, flat-range problems, the kind found in an introductory mechanics course, target archery, or a backyard trajectory check, yes, this is the standard equation and it is exact within the vacuum assumption. Real artillery, sniper ballistics, and anything crossing hundreds of metres at speed need drag coefficients, wind, and sometimes Coriolis effects layered on top, well beyond what one closed-form equation can capture.