Science & Physics FIELD NOTES · 4 MIN READ
Free Fall Explained: Why Everything Drops the Same Way
h = ½gt² and v = gt with g = 9.81 m/s² — the quadratic that makes long falls lethal and the constant that ignores mass.
On the Moon in 1971, Apollo 15 commander David Scott dropped a hammer and a falcon feather side by side. With no air to interfere, they hit the dust together. Free fall is the physics of that moment: two equations, one constant, and the cleanest predictions in all of mechanics.
The model and its two equations
Free fall means motion under gravity alone — no air resistance, no thrust. Near Earth's surface gravity accelerates every object downward at an essentially constant rate, written g, so a dropped object's height and speed are fixed functions of time:
h = v₀·t + ½·g·t² v = v₀ + g·t g = 9.81 m/s²
For a simple drop, the initial velocity v₀ is zero and the first equation collapses to h = ½·g·t². The value g = 9.81 m/s² is the internationally agreed standard gravity — measured values on Earth's surface run 9.78 to 9.83 depending on latitude and altitude, and 9.81 is the conventional middle that coursework and this site use. Every number a free-fall calculator produces flows from those two equations and nothing else.
The square in ½gt²
The half-times-t-squared term is where all the drama lives. Speed grows linearly — 9.81 m/s for every second of falling — but distance grows with the square of time, so each second of fall covers more ground than the last:
t = 1 s → h = 4.905 m v = 9.81 m/s (35.3 km/h)
t = 2 s → h = 19.62 m v = 19.62 m/s (70.6 km/h)
t = 3 s → h = 44.145 m v = 29.43 m/s (105.9 km/h)
Double the time and the distance quadruples: 2 seconds buys 19.62 m, but 4 seconds buys 78.48 m. This is why a three-storey fall (about 10 m) takes under 1.5 s — less time than a typical startle reaction — and why safety engineering treats height so aggressively. It also explains the Galilean trick of timing a well's depth: drop a stone, count the seconds until the splash, and h = ½ × 9.81 × t² gives the depth directly (with a small correction for the sound travelling back up).
g is a constant — mass is not invited
The deepest fact in free fall is what the equations leave out: mass appears nowhere. Newton's second law says a = F/m, and gravity supplies F = m × g, so the mass cancels — a = g for everything, from a dust mote to a piano. Galileo argued this from careful experiment in the 1590s, against the intuitive (and wrong) Aristotelian view that heavier falls faster; Scott's lunar hammer-and-feather was its most photogenic confirmation. Mass re-enters the story only through air resistance, because drag depends on an object's size, shape and speed while gravity depends only on its weight. In the idealised model, a bowling ball and a marble released together stay together forever.
When air resistance matters
The no-drag model is not a secret assumption — it is the definition of the tool, stated so the numbers can be judged. It is an excellent approximation for dense, compact objects over short drops, which covers most physics problems and most dropped tools. It fails as falls get long or objects get light and broad: air resistance grows with speed until it balances weight, at which point acceleration stops and the object descends at constant terminal velocity — about 55 m/s for a spread-eagle human, a few m/s for a raindrop or a ping-pong ball, and a stiff breeze for a parachute. Read the idealised numbers accordingly: without drag, computed fall times are a slight underestimate and impact speeds a slight overestimate — the real, draggy world is always gentler than the vacuum arithmetic. For converting those impact speeds between m/s, km/h and mph, speed and time units covers the factors.
Try it yourself
Open the Free Fall Calculator and enter a drop height of 19.62 m: it returns 2.00 s of fall and an impact velocity of 19.62 m/s — 70.6 km/h — with the g = 9.81, no-drag assumption on the page. Delete the height, enter 2 seconds of time instead, and the same answer arrives from the other direction: the calculator accepts either input and solves the remaining pair. Then set an initial velocity of 10 m/s with a 2-second fall to see a thrown object's extra head start: 39.62 m covered, 29.62 m/s on impact. Finally, check the quadratic against a 1-metre drop — 0.45 s — and notice how little time even a ceiling-height accident allows.