How this instrument works
Maximum height is the answer to one question: how far up does the vertical part of the motion get before gravity cancels it out? A launch at speed v and angle θ has a vertical velocity component of v·sinθ; everything else — the horizontal part, v·cosθ — carries the object sideways and never fights gravity at all. Kinematics gives the rest: at the peak the vertical velocity is exactly zero, and the identity v_y² = (v·sinθ)² − 2g·h rearranges directly into h_max = (v·sinθ)² ⁄ 2g. No calculus is needed, just the constant-acceleration equation applied at the one instant where the vertical speed vanishes.
The squared term is the reason doubling the launch speed does not double the height — it quadruples it, since (2v)² = 4v². The sinθ term is why angle matters so sharply near the extremes: sin(0°) is zero, so a perfectly horizontal launch never rises above its starting point no matter how fast it leaves, while sin(90°) is one, so a straight-up shot converts the entire speed into height and reaches the greatest peak that speed can produce, at the cost of covering no horizontal distance whatsoever.
This is also where a common mix-up lives: the 45° angle famous for maximizing RANGE is not the angle that maximizes HEIGHT. Range depends on sin(2θ), which peaks at 45°; height depends on sin²θ alone, which keeps climbing all the way to 90°. A mortar crew clearing a ridge, an artillery officer planning indirect fire over a wall, or an engineer arcing water from a fountain nozzle over an obstacle cares about this exact distinction — the angle that throws something farthest is not the angle that throws it highest.
- Enter Launch velocity — the speed the object leaves at, in m/s (switch to km/h or mph if that matches your source data).
- Set Launch angle in degrees above horizontal: 0° is dead level, 90° is straight up.
- Read Maximum height reached — the peak of the parabolic path measured above the launch point, in meters or feet.
- Hold the velocity fixed and step the angle toward 90° to see height climb while range (a separate calculation) would fall.
Worked example — 30 m/s launched at 45°
Launch velocity 30 m/s, Launch angle 45°. The vertical component is v·sinθ = 30 × sin(45°) = 21.2132 m/s. Squaring gives 450.0, and dividing by 2g = 19.6133 m/s² gives h_max = 450.0 ⁄ 19.6133 = 22.9436 m — the instrument's Maximum height reached field shows 22.943615 m for this exact pair of inputs.
Hold the speed at 30 m/s but swing the angle up to 90° and the peak rises to 45.8872 m — the tallest this launch speed can ever go, since v·sinθ now equals the full 30 m/s. Keep the angle at 45° and instead double the speed to 60 m/s and height does not double, it quadruples, to 91.7745 m, because velocity enters the formula squared while angle enters only through sine.
Questions
Why does only sinθ matter, not the full launch speed?
Because height is a purely vertical bookkeeping problem: only the vertical component of velocity, v·sinθ, ever works against gravity. The horizontal component, v·cosθ, moves the projectile sideways at a constant rate and has no effect on how high it climbs — it is entirely absent from the h_max formula for that reason.
Does a 45° launch angle give the greatest height?
No. 45° maximizes RANGE because range depends on sin(2θ), which peaks there. Height depends on sin²θ, which keeps rising all the way to 90°. At 30 m/s, 45° reaches 22.94 m while a straight-up 90° launch at the same speed reaches 45.89 m — the true maximum for that speed, with zero range.
What happens if Launch angle is set to 0°?
The height comes out as zero. sin(0°) is 0, so the vertical velocity component is 0 m/s and there is nothing to climb against gravity — a perfectly horizontal launch starts falling the instant it leaves, regardless of how fast it is moving horizontally.
Does doubling the velocity double the maximum height?
No, it quadruples it, because velocity enters the formula squared. A 30 m/s launch at 45° reaches about 22.94 m; a 60 m/s launch at the same angle reaches about 91.77 m — four times higher, not twice, for exactly double the speed.
Is Maximum height reached measured from the ground or from the launch point?
From the launch point. If the object leaves from a rooftop, a cliff edge, or a raised platform, add that starting elevation to the result separately to get height above the ground; the formula itself only tracks the rise above wherever the vertical velocity started.
Does this formula account for air resistance?
No — it assumes the textbook vacuum trajectory, with gravity as the only force acting after launch. A light, fast, or blunt object loses noticeable height to drag in real air; a dense compact one, like a shot-put or an artillery shell, tracks the formula closely over typical ranges.