How this instrument works
Drop something from rest and it falls under one force alone: gravity, pulling down at a constant 9.80665 m/s² near Earth's surface. The distance it covers grows with the square of the time elapsed, h = g t² ⁄ 2, because acceleration builds speed continuously rather than instantly. Rearranging that relationship for t is what this instrument does: t = √(2h ⁄ g) answers a narrower, more specific question than a general free-fall calculator — given only a height, how long is the object actually in the air?
The square root is the important part. Because h and t are related by a square, not a straight line, doubling the height does not double the time — it multiplies it by √2, about 1.41. A 5-metre drop takes roughly 1.01 seconds; quadruple that height to 20 metres and the time only doubles, to 2.02 seconds. That nonlinearity is exactly why a short fall feels deceptively quick and a long one deceptively slow: the seconds pile up far more slowly than the metres do.
The formula assumes two things worth stating plainly: the object starts at rest — no throw, no push, no initial speed — and nothing but gravity acts on it, which means no air resistance. Both are reasonable for a dense object dropped a modest distance, and both start to fail once the height climbs or the object is light enough for drag to matter; a dropped coin and a dropped feather share the same formula only until the air gets involved. This sheet answers the vacuum question exactly; the atmosphere is a separate calculation.
- Enter the drop height into the Height fallen field — metres for building- or cliff-scale drops, centimetres for anything smaller.
- Leave any throwing motion out of it: the formula assumes the object is simply released, starting from zero speed.
- Read the result in the Fall time field, in seconds; switch to milliseconds if the drop is short enough that seconds read as an awkward decimal.
- Compare two heights by changing Height fallen and watching how much less than proportionally Fall time moves — that square-root behavior is the formula's signature.
Worked example — a 20-metre scaffold drop
Picture a wrench slipping from a scaffold platform on the sixth floor of a building under construction, about 20 metres up. Feed h = 20 into the formula: t = √(2 × 20 ⁄ 9.80665) = √4.07887 = 2.0196 seconds. That figure is exactly what a site safety officer needs — the full window between the tool leaving a worker's hand and the tool reaching the ground below.
Two seconds sounds generous until you compare it with a barricade instead of a shout: a startled bystander typically needs three-quarters of a second just to begin reacting, leaving barely a second to actually move. Drop the same wrench from a single storey up, about 3.3 metres, and the time collapses to roughly 0.82 seconds — under the reaction threshold entirely, which is the physical reason exclusion zones sit directly beneath scaffolding rather than a few strides back.
Questions
Why does fall time follow a square root instead of scaling directly with height?
Because the distance an object falls is proportional to the square of the time, h = g t² ⁄ 2, so solving for t requires the inverse operation, a square root. Practically, that means doubling the height multiplies the fall time by √2, about 1.41, not by 2 — a 5-metre drop takes about 1.01 seconds, and reaching 20 metres, four times higher, only doubles that to 2.02 seconds rather than quadrupling it.
Does this formula still work if the object is thrown, not simply dropped?
No — it assumes zero initial velocity, an object released from rest. Throw it downward and it already has speed when the clock starts, so the true fall time is shorter than t = √(2h ⁄ g) predicts; that case needs an extra initial-velocity term this formula doesn't carry. Thrown upward or sideways, the motion splits into separate horizontal and vertical components and stops being a plain free-fall problem.
I know the impact velocity, not the height — can I still use this?
Convert it first, since this instrument only takes height as input. Impact velocity and fall time are linked by v = g t, so divide a known velocity by 9.80665 m/s² to get the time directly. To recover the height instead, use h = v² ⁄ (2g) and enter that result here — keeping the velocity and distance formulas on separate lines is usually what prevents a units mix-up.
How accurate is this for a real object, not an idealized point mass?
Exact only in a vacuum, with gravity the sole force acting. A dropped hand tool or bolt is dense and compact enough that the formula holds within a small margin over the heights typical of low-rise construction. Something with more surface area for its weight — a sheet of plywood, a scrap of insulation — feels air resistance early, stops accelerating sooner than predicted, and reaches the ground later than t = √(2h ⁄ g) says.
Why is there no mass or weight field on this instrument?
Because fall time genuinely does not depend on mass — a heavier object is pulled down harder but also resists acceleration more, in exactly matching proportion, so the two effects cancel. Galileo argued this from first principles with inclined-plane experiments centuries before anyone could drop a hammer and a feather to check it directly; the formula has one gravitational term, not two, because it needs only one.