How this instrument works
This is projectile motion with the angle fixed at zero: the object leaves moving parallel to the ground, so its initial vertical speed is exactly nothing. That single fact simplifies the general trajectory equations enormously. Vertical motion becomes plain free fall from rest, h = g t² ⁄ 2, which rearranges to t = √(2h ⁄ g) — a fall time set entirely by height and gravity, with the launch speed nowhere in the equation. Horizontal motion, meanwhile, is unopposed and constant, so distance is just speed times that same time: range = v √(2h ⁄ g).
The formula's shape comes from treating sideways and downward motion as two separate, non-interacting problems. Fire a ball horizontally off a tabletop at the same instant a second ball is simply dropped from the same edge, and the two strike the floor together, no matter how hard the first one was launched. That simultaneity is the whole content of this pair of equations: fall time answers only "how long," gravity's question; range answers only "how far," the launch speed's question, and neither ever leaks into the other's arithmetic.
Both equations assume the launch is exactly horizontal and that the landing surface sits directly below the release point, at the base of the drop — tilt the launch even slightly upward or downward and the angled trajectory equations take over instead, because a nonzero vertical launch speed changes the fall time. Air resistance is also ignored: a dense, compact object such as a ball bearing tracks these numbers closely over modest heights, but something light and wide, like a sheet of card, slows sideways too and lands well short of the predicted range.
- Enter the object's speed parallel to the ground in Launch velocity (horizontal) — the speed at the instant of release, with zero vertical component by definition.
- Enter how far above the landing surface the release point sits in Launch height, using metres or centimetres.
- Read Time to reach the ground for how long the object stays airborne — this figure moves only with height, never with speed.
- Read Horizontal range for the sideways distance covered before landing, in metres or centimetres.
- To see the two outputs behave differently, change Launch velocity (horizontal) alone: range shifts in direct proportion while Time to reach the ground does not move at all.
Worked example — a stunt car off a rooftop edge
A stunt coordinator plans a shot where a car drives straight off the flat edge of a five-metre parking structure at 72 km/h — 20 m/s — with no ramp angle, so it leaves moving purely horizontally. Feed h = 5 and v = 20 into the formulas: Time to reach the ground works out to √(2 × 5 ⁄ 9.80665) = √1.019721 = 1.0098 seconds, and Horizontal range comes to 20 × 1.0098 = 20.1962 metres. Those two numbers tell the crew exactly how long the car hangs in the air and precisely where, 20.2 metres out, the landing airbags need to sit.
The independence built into the formula is what makes the rigging predictable rather than guesswork. Swap the driver's speed to 40 m/s for a more dramatic shot and Time to reach the ground does not budge — still 1.0098 seconds, fixed by the five-metre drop alone — while Horizontal range doubles to 40.3924 metres, because range scales directly with launch speed. A coordinator who assumes a faster car means a longer hang time gets the physics backwards; only the landing zone moves, never the timing of the fall.
Questions
Why doesn't a faster horizontal launch change how long the object stays in the air?
Because gravity is the only force acting vertically, and it does not care how fast the object moves sideways. Fall time comes entirely from t = √(2h ⁄ g) — height and gravity only — so a ball rolled off a table at walking pace and one fired off the same edge at rifle speed reach the floor at the same instant, a fact classroom demonstrations confirm by releasing both together.
How is this different from the general projectile motion formula with a launch angle?
This is that formula's special case, with the launch angle fixed at zero. The general equations carry sin θ and cos θ terms and return the object to its own launch height, like a thrown ball landing where it started; here the angle term drops out entirely, the object falls to a lower surface, and range grows in direct proportion to speed rather than to speed squared.
If I double the launch velocity, what happens to the range?
It doubles too, exactly, because range = v × t and the time factor never changes with speed for a horizontal launch. That is a straight-line relationship, unlike an angled shot, where range depends on the square of speed. Halve the velocity here and the landing point moves halfway back toward the base of the drop, not to a quarter of the original distance.
Does the object's mass or weight affect the result?
No. Every mass falls under gravity at the same 9.80665 m/s², so a heavier object reaches the ground in exactly the same time as a lighter one released from the same height with the same horizontal speed — gravity pulls it down harder, but it resists that pull by exactly as much, and the two effects cancel. Neither formula on this page carries a mass term because none is needed.
What if the object is launched slightly upward or downward instead of exactly horizontal?
Then it is no longer this special case, and t = √(2h ⁄ g) will read the fall time wrong. Any vertical component to the launch, even a small one, adds to or subtracts from the time gravity needs to close the height gap, which calls for the full angled trajectory equations rather than the horizontal-only pair used here.
Can I work out the object's speed at the moment it lands, not just the time and range?
Yes, by combining the two motions Pythagorean-style, though this instrument does not output it directly. Horizontal speed stays at v throughout; vertical speed grows to g × t by landing. Impact speed is √(v² + (g t)²) — for v = 20 m/s and h = 5 m, that is √(400 + 98.07) ≈ 22.32 m/s, faster than the launch speed alone because gravity adds a downward component.