Free Fall Velocity

Equivalent forms

v^2 = u^2 + 2gh

v = gt

h = \tfrac{1}{2}gt^2

Energy conservation in one line: mgh = ½mv² → v = √(2gh).

Symbol	Name	SI unit	Dimension	Range
$v$	Final Velocityoutput Speed just before impact	m/s	LT^-1	0 – 200
$g$	Gravitational Acceleration Local gravitational acceleration	m/s²	LT^-2	0 – 25
$h$	Fall Height Vertical drop distance	m	L	0 – 1000

Unit systems

Symbol	SI	CGS	Imperial
$v$	m/s	cm/s	ft/s
$g$	m/s²	cm/s²	ft/s²
$h$	m	cm	ft

Where it holds

Valid in vacuum or for short falls where air drag is negligible. For long falls of low-density objects (raindrops, feathers), terminal velocity dominates.

Dimensional analysis

v \to [LT^{-1}]

\sqrt{g\cdot h} \to \sqrt{[LT^{-2}]\cdot [L]} = \surd [L^{2}T^{-2}] = [LT^{-1}]

Both sides are m/s.

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Discovery

Galileo Galilei · 1604

Galileo's inclined-plane experiments showed v² ∝ distance fallen, predating Newton. Torricelli applied the same formula to fluid flow from tanks.

Try this

How fast does a penny dropped from the Empire State Building actually hit the ground?

Ignore air resistance and compute v = √(2gh) for h = 380 m, g = 9.8. Then check what the real terminal velocity tells us about the myth.

Research status: stable

Real-world applications

Torricelli's law for water draining from a tank
Bungee jump landing speed calculations
Carnival drop-tower ride engineering
Building safety: object-fall impact velocity

Common misconceptions

Real falling objects do NOT keep accelerating forever — drag creates terminal velocity
v depends on h but not on the object's mass (in vacuum)
A penny from a skyscraper reaches about 25 m/s terminal, not $the \sim 85\,\mathrm{m/s}$ predicted by ignoring air

Experimental verification

Galileo's inclined-plane experiments, modern drop-tower facilities (NASA Glenn 5.2 s drop tower), and stroboscopic photography all confirm

v = \sqrt{2gh}

to high precision.

Derivation

Conservation of energy:

mgh = \tfrac{1}{2}mv^{2}

.

Cancel m and solve:

v = \sqrt{2gh}

.

Equivalently from kinematics with

u = 0

:

v^{2} = 2gh

, so

v = \sqrt{2gh}

.

Limiting cases

h \to 0

⟶

v \to 0

No drop, no impact velocity.

g \to 0

⟶

v \to 0

Zero gravity — object hovers, never accelerates.

h \to \infty

⟶

v \to \infty

(vacuum)In real atmosphere, drag caps v at terminal velocity — penny

\approx 50\,\mathrm{mph}

, not the predicted

\sim 300+ mph

.

What if…

What if h doubles?

Impact velocity scales $as \sqrt{h}$ , so $2\times$ height gives $\approx 1.41\times$ speed (not $2\times )$ . Energy doubles but velocity only grows as the square root.

What if you're on the Moon?

g_Moon $\approx 1.62\,\mathrm{m/s}^{2}$ gives $v \approx \surd (0.33\cdot v$ _Earth $^{2}) \approx 0.41\times the$ Earth speed. Moon-dropping is gentle.

What if air drag matters?

Object approaches terminal velocity $v_t = \surd (2mg/(\rho CdA))$ . Past v_t, the simple $v = \sqrt{2gh}$ over-predicts wildly.

1

Penny from the Empire State Building

Given ·

g:: 9.8
h:: 380

Find · v

Steps

Vacuum prediction: $v = \sqrt{2 \times 9.8 \times 380} = \sqrt{7448} \approx 86.3\,\mathrm{m/s}$
Real penny terminal velocity (with air drag $) \approx 25\,\mathrm{m/s}$
Conclusion: stings, doesn't kill — the myth is busted

Result · v_vacuum

\approx 86.3\,\mathrm{m/s}

; v_real

\approx 25\,\mathrm{m/s}

(drag-limited)

2

High dive from 10 m

Given ·

g:: 9.8
h:: 10

Find · v

Steps

Apply $v = \sqrt{2gh}$ with $g = 9.8$ , $h = 10$
$v = \sqrt{2 \times 9.8 \times 10} = \sqrt{196}$
$v = 14\,\mathrm{m/s}$ — Olympic platform divers hit the water $at \sim 50\,\mathrm{km/h}$

Result ·

v = \sqrt{196} = 14\,\mathrm{m/s} \approx 50\,\mathrm{km/h}

Gravitational Potential Energy→Kinetic Energy→Projectile Range→