Free fall

The most important application of constant acceleration (§2.4) in nature: free fall. Any object near Earth’s surface subject only to gravity (no air resistance) has an acceleration of g ≈ 9.8 m/s² downward — and this value is independent of mass, shape, size, or material.

This was Galileo’s discovery (~1600, see the method), and Newton later explained why: in F = mg and F = ma, mass cancels, leaving a = g — independent of m. (See gravity Article 1.)

Sign convention

Free fall is 1D (vertical). Choose a y axis:

Both are correct — just be consistent. In this section we use Convention A (up is positive). So in the kinematic equations, a = -g.

Kinematic equations for free fall

Substituting a = -g into the four equations (§2.4), the free-fall equations:

$$
v(t) = v_0 – g t
$$

$$
y(t) = y_0 + v_0 t – \frac{1}{2} g t^2
$$

$$
v^2 = v_0^2 – 2 g (y – y_0)
$$

$$
y – y_0 = \frac{v_0 + v}{2} \cdot t
$$

where v_0 is the initial vertical velocity (positive if up, negative if down) and y_0 the initial height.

Example 1 — dropping a stone

Scenario: you drop a stone from y_0 = 20 m above the ground (no initial velocity). When does it hit? What’s its impact speed?

Given: y_0 = 20, v_0 = 0, y = 0 (ground). Target: t, v.

From equation 2:

$$
0 = 20 + 0 – \frac{1}{2} \cdot 9.8 \cdot t^2
$$

$$
t^2 = \frac{40}{9.8} \approx 4.08 \Rightarrow t \approx 2.02\ \mathrm{s}
$$

From equation 1:

$$
v = 0 – 9.8 \cdot 2.02 \approx -19.8\ \mathrm{m/s}
$$

Negative sign means downward — as expected. Impact speed is |v| ≈ 19.8 m/s ≈ 71 km/h.

Quick check with equation 3:

$$
v^2 = 0 – 2 \cdot 9.8 \cdot (0 – 20) = 392 \Rightarrow |v| = \sqrt{392} \approx 19.8\ ✓
$$

Example 2 — throwing upward

Scenario: you throw a ball straight up with initial velocity v_0 = 15 m/s. What is the maximum height? When does it return to your hand?

Maximum height — at the peak, v = 0 (momentarily at rest). From equation 3:

$$
0 = 15^2 – 2 \cdot 9.8 \cdot (y_\text{max} – y_0) \Rightarrow y_\text{max} – y_0 = \frac{225}{19.6} \approx 11.5\ \mathrm{m}
$$

Time to peak — from equation 1 with v = 0:

$$
0 = 15 – 9.8 t_\text{up} \Rightarrow t_\text{up} \approx 1.53\ \mathrm{s}
$$

Time to return to hand — two ways:

Way 1: motion is symmetric. Round-trip time = 2 × up time = 2 \cdot 1.53 = 3.06 s.

Way 2: from equation 2 with y = y_0:

$$
0 = 15 t – \frac{1}{2} \cdot 9.8 t^2 = t (15 – 4.9 t)
$$

Solutions: t = 0 (throw) or t = 15/4.9 ≈ 3.06 s (return). ✓

Note: the ball returns to the starting height with the same speed but flipped sign: from +15 to -15. That symmetry is intrinsic to free fall.

Why does everything fall with the same acceleration?

From Newton’s second law:

$$
F = m a
$$

Gravitational force on an object of mass m:

$$
F = m g
$$

Hence:

$$
m a = m g \Rightarrow a = g
$$

Mass cancels. A feather and a hammer (with air resistance removed) fall together. That famous experiment was carried out by Apollo 15 astronaut David Scott on the Moon (no atmosphere) — feather and hammer hit the surface at exactly the same instant.

This has a deeper consequence: gravitational mass equals inertial mass — a fact not explained by Newtonian physics but by general relativity (Einstein’s equivalence principle).

Air resistance — where the ideal breaks down

Free-fall assumes no air resistance. Reality:

Rule of thumb: for heights under ~50 m and dense objects, air resistance is at most a few-percent error. For light objects or larger heights, you must add a drag model (F_drag ∝ v² or v) — and it’s no longer constant acceleration.

A few notes and common mistakes

1. g is not a universal constant — it’s a local value.
g ≈ 9.8 m/s² is a surface approximation. On Everest ~9.78, in the ISS ~8.7, on the Moon ~1.62. (See gravity Article 3.) In most problems, 9.8 or 10 is used — but don’t treat it as a “law”.

2. Is acceleration “zero at the peak”?
No. At the peak, velocity is zero but acceleration is still -g. That’s why the ball immediately starts moving downward. If acceleration were zero too, the ball would hang in the air.

3. Initial velocity doesn’t have to be positive.
If you throw up, v_0 > 0. If you throw down, v_0 < 0. If you drop, v_0 = 0. Read the sign from the initial motion.

4. Round-trip symmetry.
If you launch and return to the same height, impact speed = launch speed (only the sign flips). If it ends at a different height, use equation 3 to compute the final speed.

5. Free fall from very high altitude is not ideal.
From 100 km altitude (like the Redbull capsule), g is still ~9.5 but air resistance dominates. Falling “freely” from space is a misconception — there’s always a terminal velocity as long as there’s atmosphere.

What you should be able to do

After this section, you should be able to:

Preview of §2.6

So far we’ve only worked with constant acceleration. But if acceleration varies (air drag, springs, variable-power engine), the §2.4 kinematic equations don’t apply. We have to go back to the primary definitions and use graphical analysis + integration. §2.6 covers:

which are the inverses of differentiation. These give a powerful method that works for any a(t) — not just constant.

📚 See also: Halliday Vol 1, Ch 2, §2.5 — Free-Fall Acceleration.
📖 Open reference: OpenStax University Physics Vol 1 — Chapter 3.5: Free Fall.
📖 The feather-and-hammer simultaneity test on the Moon: NASA Apollo 15 Feather and Hammer.

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