Free fall — g = 9.8 m - s², sign conventions: Visual walkthrough

Step 1 — Draw the number line first, before any physics

WHAT. Before we talk about falling, we draw a vertical ruler. We choose up = positive. The ground and the sky are now just numbers: a spot 5 metres up is $+ 5$ , a spot 3 metres below the start is $- 3$ .

WHY. Every " $+$ " and " $-$ " in free fall is a choice we make once. If we don't pin down the ruler now, later the signs feel like magic. They are not magic — they are consequences of this one arrow.

PICTURE. In the figure, the purple arrow shows the direction we called positive. Everything pointing with it counts as $+$ ; everything pointing against it counts as $-$ . Notice gravity's little coral arrow points against our positive direction. Hold that thought — it is the entire reason $g$ will wear a minus sign.

Figure — Free fall — g = 9.8 m - s², sign conventions

Step 2 — What "acceleration" means: velocity's slope

WHAT. We claim free fall has constant acceleration: velocity changes by the same amount every second. Near Earth that amount has size $g \approx 9.8 m/s^{2}$ , directed down. On our up-positive ruler that makes $a = - g = - 9.8 m/s^{2} .$

WHY this number, why constant? From Newton's Second Law: gravity pulls with force $m g$ , and $F = ma$ gives $ma = m g$ , so $a = g$ — the mass cancels, and nothing in that equation depends on time, so $a$ never changes. One value, forever (until something hits the ground or air joins in — see Air Resistance and Terminal Velocity).

PICTURE. Plot velocity $v$ against time $t$ . Because $a$ is constant, this graph is a straight line. Its steepness is the acceleration. Below we let $\frac{d v}{d t}$ mean exactly "steepness of the $v$ – $t$ line" — the amount $v$ gains per second.

$a = steepness of the v - t line \frac{d v}{d t} = - g$

Step 3 — From slope back to velocity: the first equation

WHAT. We want a formula for $v$ at any time $t$ . We know the line starts at height $v_{0}$ (the initial velocity) and has slope $a = - g$ . A straight line is start value + slope × run: $v = start height of line v_{0} + slope a run along time t = v_{0} - g t .$

WHY. This is just reading the straight line off the graph — no calculus needed once we see it's a line. (The parent's integral $\int d v = \int a d t$ says the identical thing: adding up equal little velocity-gains gives slope × time.)

PICTURE. Follow the coral line down. Every second it drops by $9.8$ . It starts positive (thrown up), crosses zero (the peak — the ball is momentarily still), then goes negative (falling back). One straight line contains the whole up-and-down story.

Step 4 — From velocity to position: area under the line

WHAT. Position changes because velocity moves us along the ruler. The displacement (how far $y$ moved) equals the area between the $v$ – $t$ line and the time axis.

WHY area? Distance = speed × time. If speed were constant, that's a rectangle. Here speed changes along a straight line, so the region is a rectangle (from $v_{0}$ ) plus a triangle (from the changing part). Area of shapes we know — that is why the tool here is "area," not something fancier.

PICTURE. The mint rectangle has height $v_{0}$ and width $t$ : area $v_{0} t$ . The lavender triangle has width $t$ and height $g t$ (how much velocity dropped): area $\frac{1}{2} (t) (g t) = \frac{1}{2} g t^{2}$ . Because the triangle sits below the axis on the way down, it subtracts.

$y - y_{0} = rectangle v_{0} t - triangle \frac{1}{2} g t^{2}$

Step 5 — Killing time: the speed–distance relation

WHAT. Sometimes we don't care about $t$ , only about how fast after how far. We remove $t$ from the two boxed equations.

WHY. Problems like "impact speed after a 20 m drop" give distance, not time. A formula linking $v$ and displacement $Δ y$ directly saves us solving for $t$ first.

HOW (the algebra, in words). From Equation 1, $t = \frac{v _{0} - v}{g}$ . Feed that into Equation 2 and the $t$ 's all collapse. The clean result: $v^{2} = v_{0}^{2} - 2 g Δ y, Δ y = y - y_{0} .$

PICTURE. Think of it as an energy ledger drawn on the ruler: rising by $Δ y$ costs $2 g Δ y$ of " $v^{2}$ "; falling refunds it. Height and speed-squared trade one-for-one.

Step 6 — The peak, and why acceleration is not zero there

WHAT. Set $v = 0$ in Equation 1: the object is momentarily still. Solve $0 = v_{0} - g t$ to get the time to the top: $t_{up} = \frac{v _{0}}{g} .$ Plug into Equation 3 with $v = 0$ : the max height $H = \frac{v _{0}^{2}}{2 g} .$

WHY it matters. Beginners think "stopped = no acceleration." But on the $v$ – $t$ line, the peak is just the single point where the line crosses zero — the line's slope there is still $- g$ . Velocity passes through zero; acceleration sails right through unchanged.

PICTURE. The coral line touches the time-axis at $t_{up}$ (velocity zero) but keeps its downward slant. The height curve $y (t)$ reaches its rounded summit at that same instant — flat there, but curving down the whole time.

Step 7 — All the cases on one picture

WHAT. The same three equations cover every launch. Only the sign of $v_{0}$ and the starting height change.

WHY show them together. So you never meet a scenario the derivation skipped:

Case	$v_{0}$	What the line does
Dropped from rest	$0$	starts on the axis, only falls
Thrown up	$+$	starts positive, crosses zero at the peak, then negative
Thrown down	$-$	starts negative, gets more negative

PICTURE. Three velocity lines, same slope $- g$ (all parallel!), different starting heights $v_{0}$ . Parallel = same acceleration for every object and every throw — the deep truth of free fall in one glance. And the degenerate case (dropped, $v_{0} = 0$ ) is simply the line that starts on the axis.

The one-picture summary

WHAT. One straight $v$ – $t$ line generates everything: its starting height is $v_{0}$ , its slope is $- g$ (Equation 1), its area is displacement (Equation 2), and its zero-crossing is the peak. Kill $t$ and you get Equation 3. That's the whole chapter compressed into a line and the region under it.

Recall Feynman retelling — the whole walkthrough in plain words

First I drew a ruler standing up and said "up is the plus direction." Gravity points the other way, so it always shows up as a minus. Then I noticed that gravity speeds a falling thing up by the same 9.8 every single second — so if I plot how fast it's going against time, I get a perfectly straight, tilted line. A straight line is "where you started plus your slope times the time," and that's my first equation, $v = v_{0} - g t$ . To find where the thing is, I measured the area trapped under that line — a rectangle plus a triangle — and out popped $y = y_{0} + v_{0} t - \frac{1}{2} g t^{2}$ . When I only cared about speed after some distance and not the clock, I algebraically deleted the time and got $v^{2} = v_{0}^{2} - 2 g Δ y$ . Setting the speed to zero found the top of the throw — but the line's tilt never flattened there, which is exactly why gravity keeps pulling and the thing comes back down. Finally I drew three throws at once: dropped, up, down — three parallel lines, same tilt — a picture of the fact that everything falls the same.

Active recall

Connections

Equations of Motion (constant acceleration) — this page is the $a = - g$ special case, drawn out
Vectors and sign conventions — the ruler in Step 1
Newton's Second Law — why $a = g$ with mass cancelled (Step 2)
Projectile Motion — attach a steady sideways motion and this becomes the vertical half
Air Resistance and Terminal Velocity — what bends the straight $v$ – $t$ line
Parent: Free fall — g = 9.8 m - s², sign conventions (index 1.1.17)