Free Fall

Square the fall time, multiply by g = 9.81 m/s² and halve the result. A stone falling for t = 2 s covers h = ½ × 9.81 × 4 = 19.62 m, roughly a six-storey building. The formula holds for a fall from rest without air resistance. Useful rules of thumb: after 1 s about 5 m, after 2 s about 20 m, after 3 s about 45 m; the heights grow quadratically. The impact speed follows separately from v = g·t or v = √(2gh). In many school problems g ≈ 10 m/s² may be used; check what the task specifies.

In vacuum all bodies fall exactly equally fast; the mass cancels: from F = m·g and F = m·a follows a = g for every mass. David Scott demonstrated this impressively on the Moon in 1971: hammer and feather landed simultaneously. In everyday life it seems otherwise because air resistance interferes: it depends on shape and cross-sectional area, not on mass. A sheet of paper glides because the air force is large relative to its small weight; crumpled up, it falls almost like a stone. The formula h = ½gt² describes idealised free fall; for compact, heavy objects over a few metres it is an excellent approximation.

Rearrange h = ½gt² for the time: t = √(2h/g). Multiply the height by 2, divide by 9.81 and take the square root. Example: from a 45 m cliff the fall takes t = √(90/9.81) = √9.17 ≈ 3.0 s. Because of the root, the fall time grows only slowly with height: for four times the height a body needs merely twice the time. The well-known well trick uses this formula: drop a stone, count the seconds until the splash, estimate the depth with h ≈ 5·t². For precise calculations at large depths you would additionally account for the sound travelling back up, a popular extension in exam problems.

Two equivalent routes lead to the answer. If you know the fall time, v = g·t applies: after 2 s that is 9.81 × 2 = 19.62 m/s, about 71 km/h. If you know the height, use v = √(2gh), which follows directly from energy conservation m·g·h = ½·m·v²: from 20 m you get v = √(2 × 9.81 × 20) ≈ 19.8 m/s. Note that the mass appears in neither formula. A common mistake is confusing the average speed h/t with the final speed; in free fall from rest the final speed is exactly twice the average, because v grows linearly with time.

If a body is thrown straight down, the initial term is added: h = ½gt² + v₀t; the structure is the same as the general distance-time law with a = g. For a vertical throw upwards you set v₀ against g; at the highest point v = 0. On other celestial bodies you only replace the local factor: Moon g ≈ 1.62 m/s², Mars g ≈ 3.71 m/s², Jupiter g ≈ 24.8 m/s². The same stone falling 20 m on the Moon takes t = √(2·20/1.62) ≈ 5 s instead of 2 s on Earth. The formulas stay identical; only the constant g is planet-specific. Exactly such transfer questions are popular in final exams.