How this instrument works
Horizontal range is how far a projectile travels before it drops back to the height it launched from — not its peak, not how long it stayed up, just the flat distance between launch and landing. The formula comes from treating the flight as two independent problems running on the same clock. The horizontal component of the launch velocity, v·cosθ, never changes because nothing pushes back on it; the vertical component, v·sinθ, is erased by gravity on the way up and rebuilt on the way down, so the object spends 2v·sinθ ⁄ g seconds aloft. Range is that constant horizontal speed multiplied by the total time in the air, and the double-angle identity sin(2θ) = 2 sinθ cosθ folds the two velocity components into the single compact form R = v²sin(2θ) ⁄ g.
The 45° maximum is a compromise between two effects pulling in opposite directions, not a rule about throwing harder. Horizontal reach favors a flat angle, since cosθ is largest near 0°; hang time favors a steep one, since sinθ is largest near 90°. Range depends on both at once, through the product sinθ·cosθ, and that product is greatest exactly where the two competing terms balance — at θ = 45°, where sinθ and cosθ are equal. Move the angle either direction from there and one factor grows while the other shrinks faster, which is why range falls away symmetrically on both sides of the peak rather than sloping off in only one direction.
The formula only holds when the launch point and the landing point sit at the same height and nothing but gravity acts on the object in between. A ball rolled off a raised table, a stone thrown from a cliff edge toward the ground below, or a target mounted higher than the throwing hand all break that same-height assumption and need the fuller trajectory equation that tracks height separately. It is also worth remembering that range alone never describes a flight path: a flat, fast arc and a high, slow one can land at the exact same distance while climbing to completely different peak heights and staying airborne for completely different times, so this single number is deliberately blind to everything except where the object comes down.
- Enter Launch velocity — the speed the object leaves at, in m/s or mph.
- Enter Launch angle — the angle above horizontal, in degrees. The default, 45°, is the angle that maximizes range for any Launch velocity.
- Read Horizontal range — the flat-ground distance covered before the object returns to launch height, in meters or feet.
- Hold Launch velocity fixed and step Launch angle away from 45° in either direction to see Horizontal range fall off symmetrically.
Worked example — testing the 45° range maximum with a lab launcher
An introductory physics lab sets a spring-loaded ball launcher to Launch velocity 20 m/s and Launch angle 45° (0.785398 radians) to test the textbook claim that this angle throws farthest on flat ground. Because sin(2 × 45°) = sin(90°) = 1 exactly, the formula collapses to v² ⁄ g: 20² ⁄ 9.80665 = 400 ⁄ 9.80665 = 40.7886485191 meters, which the Horizontal range field reports as 40.788649 m — the largest distance that launcher can reach at that speed, whatever angle is dialed in.
Leaving Launch velocity at 20 m/s and resetting Launch angle to 30° drops Horizontal range to 35.3240058036 meters, and resetting it instead to 60° returns the identical 35.3240058036 meters, because sin(60°) and sin(120°) share the same value. Only 45° sits at the top of that curve. Raising Launch velocity to 40 m/s at 45° does not simply double the earlier result — it quadruples it, to 163.154594076 meters, since the formula carries velocity squared rather than velocity alone.
Questions
Why does a 45° launch angle produce the maximum range?
Because range is proportional to sin(2θ), and sine reaches its largest possible value, 1, only when its argument is 90° — which happens exactly when θ = 45°. Any other angle makes sin(2θ) smaller than 1, so the same Launch velocity produces a shorter Horizontal range; it is a property of the sine function itself, not of gravity or the object being thrown.
Do two different launch angles ever give the same range?
Yes — any pair of angles that add up to 90°, such as 30° and 60° or 20° and 70°, gives an identical Horizontal range for the same Launch velocity, because sin(2θ) and sin(2(90° − θ)) are always equal. At 20 m/s, both 30° and 60° reach 35.3240058036 meters, though one path climbs higher and stays up longer than the other.
Does this formula also give the maximum height or flight time?
No — Horizontal range is the only quantity this formula produces. Peak height and time aloft come from separate expressions built on the same v·sinθ and v·cosθ components, so a Launch velocity and Launch angle that fix the range still leave the height and flight-time questions open; a different calculation is needed for either one.
What if the target is higher or lower than the launch point?
Then this formula does not apply directly, since it assumes the object lands at the same height it left from. A target above or below launch height needs the fuller trajectory equation that keeps a separate height term, because the climb and the fall stop being mirror images of each other once takeoff and landing heights differ.
Why does doubling the launch velocity more than double the range?
Because velocity enters the formula squared while the angle only ever contributes a multiplier between 0 and 1. Raising Launch velocity from 20 m/s to 40 m/s at the same 45° angle takes Horizontal range from 40.7886485191 meters to 163.154594076 meters — four times as far, not two, since (2v)² equals 4v².
What units does this calculator accept and return?
Launch velocity accepts m/s or mph, Launch angle is entered in degrees, and Horizontal range can be read in meters or feet from its own unit menu. Switching any of these menus does not change the underlying arithmetic — internally everything is converted to meters, seconds, and radians before the formula runs — it only changes how the numbers are displayed.