Choose what you know — the height, the fall time or the impact speed — and pick the gravity. The object is assumed to be dropped from rest with no air resistance.
An object dropped from rest speeds up at a constant rate g (9.81 m/s² on Earth). The three free-fall equations link height h, time t and final speed v: h = ½gt², v = gt, and v = √(2gh). Knowing any one of them gives the other two.
Dropping something from 20 m takes t = √(2h ÷ g) = √(40 ÷ 9.81) ≈ 2.02 s, and it hits the ground at about 19.8 m/s (71 km/h or 44 mph). On the Moon the same drop takes almost 5 seconds and lands at 8 m/s.
In real life, air resistance matters: a skydiver stops accelerating at roughly 55 m/s (200 km/h) terminal velocity. These results are the idealised, vacuum values used in physics classes.
About 1.43 seconds on Earth, using t = √(2h ÷ g) = √(20 ÷ 9.81). The object would hit the ground at about 14 m/s (50 km/h), ignoring air resistance.
Not in a vacuum: all objects fall with the same acceleration regardless of mass. In air, heavier and more streamlined objects are slowed less by drag, which is why a feather falls slower than a hammer on Earth but not on the Moon.
The acceleration due to gravity at the surface: about 9.81 m/s² (32.2 ft/s²) on Earth. It varies slightly with latitude and altitude, and is much lower on the Moon (1.62 m/s²).
Use h = ½gt². A stone that takes 3 seconds to hit the water fell about ½ × 9.81 × 9 = 44 m, if you ignore the time for the sound to come back up.
No. It gives the ideal free-fall values. Real objects, especially light or wide ones, fall more slowly and eventually stop speeding up at their terminal velocity.
