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Physics · Mechanics

Free Fall

Drop something from rest and get the fall time and the speed it hits the ground.

Known value
m
The distance the object falls, in metres.
Fall time
2.02s
Impact speed
19.81m/s
Distance fallen over time (accelerating)
Distance fallen grows as ½ × g × t², reaching 20 m after 2.02 s of free fall20 m0 m0 s2.02 s

An object dropped from rest falls with time t = √(2h ÷ g) and lands at speed v = √(2gh), where g = 9.81 m/s². Drop it from 20 m and it falls for about 2.02 s, hitting the ground at roughly 19.8 m/s (about 71 km/h).

What free fall means

Free fall is motion under gravity alone, with no air resistance and no initial push. The object starts from rest, so its only acceleration is g ≈ 9.81 m/s² downward. Because that acceleration is constant, the same two relationships always hold: the fall time depends only on the height, and the impact speed depends only on the height (or, equivalently, on g times the time). This is a special case of the SUVAT equations of motion with initial velocity u = 0.

h = ½gt²  ·  t = √(2h ÷ g)  ·  v = √(2gh) = g·t

h = drop height, t = fall time, v = impact speed, g = 9.81 m/s² (drop from rest, no air resistance)

Worked example

Drop an object from a height of h = 20 m with g = 9.81 m/s²:

  1. 1
    Find the fall time. t = √(2h ÷ g) = √(2 × 20 ÷ 9.81) = √4.077 ≈ 2.02 s.
  2. 2
    Find the impact speed from the height. v = √(2gh) = √(2 × 9.81 × 20) = √392.4 ≈ 19.81 m/s.
  3. 3
    Cross-check with v = g·t. v = g × t = 9.81 × 2.02 ≈ 19.81 m/s — the two routes agree.

Fall time and impact speed by drop height

Dropped from rest, g = 9.81 m/s², air resistance ignored.

HeightFall timeImpact speed
1 m0.45 s4.43 m/s (16 km/h)
5 m1.01 s9.90 m/s (36 km/h)
10 m1.43 s14.01 m/s (50 km/h)
20 m2.02 s19.81 m/s (71 km/h)
50 m3.19 s31.32 m/s (113 km/h)
100 m4.52 s44.29 m/s (159 km/h)

Assumptions and limits

Air resistance is ignored. These formulas assume a vacuum. In real air, drag grows with speed and eventually limits the fall — a feather and a hammer only land together on the Moon.

Mass does not matter. The mass cancels out, so a bowling ball and a marble dropped together (in a vacuum) reach the ground at the same instant.

g varies slightly. We use g = 9.81 m/s², the standard value near Earth’s surface; it ranges from about 9.78 at the equator to 9.83 at the poles, and is far smaller on the Moon (≈ 1.62 m/s²).

Does the mass of the object affect how fast it falls?
No. In free fall with no air resistance, every object accelerates at the same g ≈ 9.81 m/s² regardless of mass. A heavy ball and a light ball dropped from the same height land at the same time.
Why does mass cancel out?
Newton’s second law gives a = F ÷ m, and the gravitational force is F = m·g. The mass on top cancels the mass on the bottom, leaving a = g — so acceleration is independent of mass.
Does this calculator account for air resistance?
No. It assumes free fall in a vacuum, so the results are exact only when drag is negligible — short drops, dense compact objects. For a feather or a parachute, real fall times are much longer.
What is terminal velocity and why isn’t it here?
Terminal velocity is the steady speed where air drag balances gravity, so the object stops accelerating. It depends on shape, mass, and air density, which this idealised model ignores — real falls level off near it (a skydiver reaches roughly 55 m/s).
What value of g does this use?
It uses g = 9.81 m/s², the standard gravitational acceleration near Earth’s surface. The true value varies slightly with latitude and altitude (≈ 9.78–9.83 m/s²) and is about 1.62 m/s² on the Moon.
How do I find the impact speed from the time instead?
Use v = g·t. With g = 9.81 m/s², an object falling for 2.02 s hits the ground at 9.81 × 2.02 ≈ 19.81 m/s, the same as √(2gh) for a 20 m drop.