Free fall — g = 9.8 m - s², sign conventions: Worked examples

Before anything, let me re-anchor the three tools so you never have to scroll away.

Here, $Δ y = y - y_{0}$ means "final position minus starting position" — the displacement, not the distance travelled. That distinction matters and we will see it bite.

The scenario matrix

Every free-fall problem is really "which of these cells am I in?" Each cell differs in the sign of $v_{0}$ , whether the landing point is above/below/at the start, and whether some input is zero or pushed to a limit.

#	Cell class	What's special	Example
1	Pure drop, $v_{0} = 0$	starting velocity is zero (a degenerate input)	Ex 1
2	Thrown up, lands back at start	symmetric flight, $Δ y = 0$	Ex 2
3	Thrown up, lands below start	$Δ y < 0$ ; the ball passes its launch height going down	Ex 3
4	Thrown down, both signs positive	"down positive" convention keeps everything $+$	Ex 4
5	Same problem, opposite convention	prove the answer doesn't depend on choice	Ex 5
6	Quadratic with two roots	which time do we keep? sign of the root	Ex 6
7	Limiting value	what happens as $v_{0} \to 0$ or $g \to$ small	Ex 7
8	Real-world word problem	reaction time, a moving reference	Ex 8
9	Exam twist	two objects, or a hidden assumption	Ex 9

We now hit each cell in order.

Ex 1 — Cell 1: pure drop ( $v_{0} = 0$ )

Setup (why?): "Released from rest" means the degenerate input $v_{0} = 0$ — no throw, gravity does all the work. Choose up positive, put the ground at $y = 0$ , so $y_{0} = + 45$ , $a = - 9.8$ .

Find the time. Use equation (2) with $y = 0$ . $0 = 45 + 0 \cdot t - \frac{1}{2} (9.8) t^{2} .$ Why this step? We want the instant the ball is at the ground, i.e. $y = 0$ . The $v_{0} t$ term vanishes because $v_{0} = 0$ — notice the degenerate input simplifies the equation rather than breaking it.
Solve for $t$ . $4.9 t^{2} = 45 ⟹ t^{2} = 9.184 ⟹ t = 3.03 s .$ Why this step? Simple algebra; only the positive root makes physical sense (negative time is before release).
Find impact velocity. Use equation (1). $v = 0 - 9.8 (3.03) = - 29.7 m/s .$ Why this step? The sign tells direction — negative = downward in our convention. Speed is $29.7 m/s$ .

Verify: Cross-check with the no-time equation (3): $v^{2} = 0 + 2 (9.8) (45) = 882$ , so $∣ v ∣ = 882 = 29.7 m/s$ . ✓ Matches. Units: $(m/s^{2}) (m) = m^{2} / s^{2} = m/s$ . ✓

Ex 2 — Cell 2: thrown up, returns to start ( $Δ y = 0$ )

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

Setup: Up positive, $y_{0} = 0$ , $v_{0} = + 12$ , $a = - 9.8$ .

Max height. At the peak the ball momentarily stops, so $v = 0$ . Use equation (3): $0 = 1 2^{2} - 2 (9.8) H ⟹ H = \frac{144}{19.6} = 7.35 m .$ Why $v = 0$ ? Look at the figure: velocity shrinks as the ball rises, hits exactly zero at the apex (the plum dot), then grows downward. The instant of turnaround is where $v = 0$ .
Time to top. From equation (1): $0 = 12 - 9.8 t ⟹ t_{up} = 1.22 s$ . Why this step? Same " $v = 0$ at top" idea, now solved for time.
Total flight time. By up–down symmetry (the parabola in the figure is a mirror image about the apex), $t_{total} = 2 t_{up} = 2.45 s$ . Why this step? Because $Δ y = 0$ over the whole trip, equation (2) gives $0 = 12 t - 4.9 t^{2} = t (12 - 4.9 t)$ : roots $t = 0$ (launch) and $t = 2.45$ (return). The two roots are the two times the ball is at hand height.

Verify: Return speed from (1): $v = 12 - 9.8 (2.45) = - 12 m/s$ — same magnitude, now downward. ✓ Exactly the symmetry we predicted.

Ex 3 — Cell 3: thrown up, lands below start ( $Δ y < 0$ )

Setup: Up positive, launch point $y_{0} = 0$ , ground at $y = - 25$ (below start ⟹ negative), $v_{0} = + 8$ , $a = - 9.8$ . Here the displacement $Δ y = - 25 - 0 = - 25 m$ is negative — this is the cell people fumble.

Impact speed via (3). $v^{2} = 8^{2} + 2 (- 9.8) (- 25) = 64 + 490 = 554.$ Why this step? Equation (3) skips the time and the up-then-down detour — perfect when we only want speed. Note the two negatives multiply to a positive, so the drop adds speed. $∣ v ∣ = 554 = 23.5 m/s (downward, so v = - 23.5) .$
Impact time via (2). $- 25 = 8 t - 4.9 t^{2} ⟹ 4.9 t^{2} - 8 t - 25 = 0.$ Why this step? Now we need time, so we use position (2). This is a quadratic; solve it: $t = \frac{8 \pm 64 + 4 ( 4.9 ) ( 25 )}{2 ( 4.9 )} = \frac{8 \pm 554}{9.8} .$
Pick the root. $554 = 23.5$ , so $t = \frac{8 + 23.5}{9.8} = 3.22 s$ or $t = \frac{8 - 23.5}{9.8} = - 1.58 s$ . Why keep the first? Negative time is before the throw — physically meaningless here. Keep $t = 3.22 s$ .

Verify: Feed $t = 3.22$ into (1): $v = 8 - 9.8 (3.22) = - 23.6 m/s$ (rounding). ✓ Magnitude matches step 1. And impact speed ( $23.5$ ) exceeds launch speed ( $8$ ) — sensible, the cliff gave the ball extra fall distance.

Ex 4 — Cell 4: thrown down, all-positive convention

Setup (why this convention?): Everything moves down, so choose down positive — then $v_{0} = + 6$ , $a = + 9.8$ , and every number stays positive, killing sign slips. Start $y_{0} = 0$ .

Depth via (2). $y = 0 + 6 (2.0) + \frac{1}{2} (9.8) (2.0)^{2} = 12 + 19.6 = 31.6 m .$ Why this step? We know the time and want the position — that is exactly what (2) delivers.
Impact speed via (1). $v = 6 + 9.8 (2.0) = 25.6 m/s .$ Why this step? Velocity after a known time — equation (1). Positive means downward here (as intended).

Verify: Cross-check with (3): $v^{2} = 6^{2} + 2 (9.8) (31.6) = 36 + 619.4 = 655.4$ , $655.4 = 25.6 m/s$ . ✓ Depth $31.6 > 19.6$ , so the throw added depth — sensible.

Ex 5 — Cell 5: SAME problem, up-positive convention

Setup: Up positive. Thrown down so $v_{0} = - 6$ . Gravity down so $a = - 9.8$ . Water is below start, so its position will be $y = - 31.6$ (negative, since below). $y_{0} = 0$ .

Position at $t = 2.0$ via (2). $y = 0 + (- 6) (2.0) + \frac{1}{2} (- 9.8) (2.0)^{2} = - 12 - 19.6 = - 31.6 m .$ Why this step? Same equation, but now both $v_{0}$ and $a$ are negative. Depth = $∣ - 31.6 ∣ = 31.6 m$ .
Velocity at $t = 2.0$ via (1). $v = - 6 + (- 9.8) (2.0) = - 25.6 m/s .$ Why this step? Negative because it's downward in this convention. Speed = $25.6 m/s$ .

Verify: Depth $31.6 m$ and speed $25.6 m/s$ — identical to Ex 4. ✓ The physics never depends on which way you call positive; only the signs flip. This is the whole point of the Vectors and sign conventions discipline.

Ex 6 — Cell 6: quadratic with two meaningful roots

Setup: Up positive, $y_{0} = 0$ , $v_{0} = + 20$ , $a = - 9.8$ , target $y = 15$ .

Write equation (2) at $y = 15$ . $15 = 20 t - 4.9 t^{2} ⟹ 4.9 t^{2} - 20 t + 15 = 0.$ Why this step? We fix the height and ask "for which $t$ ?" — a quadratic, so up to two solutions.
Solve. $t = \frac{20 \pm 400 - 4 ( 4.9 ) ( 15 )}{2 ( 4.9 )} = \frac{20 \pm 106}{9.8} .$ $106 = 10.30$ , so $t = \frac{20 + 10.30}{9.8} = 3.09 s$ or $t = \frac{20 - 10.30}{9.8} = 0.99 s$ .
Interpret BOTH roots. Look at the figure: the ball passes $15 m$ once going up ( $t = 0.99 s$ , orange dot) and once coming down ( $t = 3.09 s$ , teal dot). Both are physical — unlike Ex 3, we keep them both. Why keep both? The discriminant is positive and both roots are positive times after launch. The height $15$ is below the apex ( $H = 400/19.6 = 20.4 m$ ), so the ball reaches it twice.

Verify: If the two roots straddle the apex, their average should equal the time-to-top $t_{up} = 20/9.8 = 2.04 s$ . Average $= \frac{0.99 + 3.09}{2} = 2.04 s$ . ✓ Beautiful symmetry check.

Ex 7 — Cell 7: the limiting case ( $v_{0} \to 0$ )

Setup: Use $H = v_{0}^{2} / (2 g)$ with $g = 9.8$ .

Plug in the three speeds. $H (10) = \frac{100}{19.6} = 5.10 m, H (1) = \frac{1}{19.6} = 0.051 m, H (0.1) = \frac{0.01}{19.6} = 0.00051 m .$ Why this step? Watching the value shrink shows the shape of the dependence.
Read off the pattern. Divide $v_{0}$ by 10 and $H$ drops by $100$ — because $H \propto v_{0}^{2}$ , a quadratic dependence, not linear. Why does this matter? It tells us the sensitivity: near zero, a small throw barely lifts the ball.
Take the limit. $lim_{v_{0} \to 0} H = lim_{v_{0} \to 0} \frac{v _{0}^{2}}{2 g} = 0.$ Why this step? This is the sanity boundary between "throw up" (Cells 2–3) and "pure drop" (Cell 1): a throw of zero speed is a drop, and its rise height is exactly $0$ . The formulas connect seamlessly — no discontinuity.

Verify: Does $v_{0} = 0$ in the drop of Ex 1 rise at all? No — it starts falling immediately, $H = 0$ . ✓ Consistent with the limit. The two cells meet exactly at $v_{0} = 0$ .

Ex 8 — Cell 8: real-world word problem (reaction time)

Setup: Down positive (only falling). $v_{0} = 0$ (released from rest), $a = + 9.8$ , $d = 0.18 m$ .

Use (2) to link distance and time. $d = \frac{1}{2} g t^{2} ⟹ 0.18 = \frac{1}{2} (9.8) t^{2} .$ Why this step? We know the fallen distance and want the elapsed time — position equation, with $v_{0} = 0$ (a Cell-1 style degenerate input embedded in a word problem).
Solve for $t$ . $t^{2} = \frac{0.36}{9.8} = 0.03673 ⟹ t = 0.192 s .$ Why this step? Positive root only — time can't be negative. Reaction time $\approx 0.19 s$ .

Verify: Impact speed if useful: $v = 2 (9.8) (0.18) = 1.88 m/s$ . The time $0.19 s$ sits right in the human reaction-time band ( $0.15$ – $0.25 s$ ). ✓ Physically believable — this is literally how the "ruler drop test" works.

Ex 9 — Cell 9: exam twist (two objects)

Setup: Down positive, $y_{0} = 0$ , ground at $y = + 60$ . Ball A: $v_{0} = 0$ . Ball B: $v_{0} = + 10$ . Both $a = + 9.8$ .

Time for A (a pure drop, Cell 1). $60 = \frac{1}{2} (9.8) t_{A}^{2} ⟹ t_{A}^{2} = \frac{120}{9.8} = 12.24 ⟹ t_{A} = 3.50 s .$ Why this step? No initial velocity, so only the $\frac{1}{2} a t^{2}$ term survives.
Time for B (Cell 4 style). $60 = 10 t_{B} + \frac{1}{2} (9.8) t_{B}^{2} ⟹ 4.9 t_{B}^{2} + 10 t_{B} - 60 = 0.$ $t_{B} = \frac{- 10 + 100 + 4 ( 4.9 ) ( 60 )}{2 ( 4.9 )} = \frac{- 10 + 1276}{9.8} = \frac{- 10 + 35.72}{9.8} = 2.62 s .$ Why keep the $+$ root? The $-$ root gives negative time (before launch), discarded.
Difference. $Δ t = t_{A} - t_{B} = 3.50 - 2.62 = 0.88 s$ . Why this step? The question asks for the gap between impacts.

Verify: Sanity: B has a head start of speed, so it must land first ( $t_{B} < t_{A}$ ) — ✓. Check A's impact speed $v_{A} = 2 (9.8) (60) = 34.3 m/s$ vs B's $v_{B} = 1 0^{2} + 2 (9.8) (60) = 1276 = 35.7 m/s$ ; B is faster, consistent with landing sooner. ✓

Recap: which cell was which

Recall Map each example back to its matrix cell

Ex 1 → Cell 1 (drop, $v_{0} = 0$ ) · Ex 2 → Cell 2 ( $Δ y = 0$ symmetric) · Ex 3 → Cell 3 ( $Δ y < 0$ ) · Ex 4 → Cell 4 (down positive) · Ex 5 → Cell 5 (convention swap) · Ex 6 → Cell 6 (two roots) · Ex 7 → Cell 7 (limit $v_{0} \to 0$ ) · Ex 8 → Cell 8 (word problem) · Ex 9 → Cell 9 (two objects).

Active recall

Which equation skips time entirely?

v^{2} = v_{0}^{2} + 2 a Δ y

(equation 3).

In Ex 3, why did impact speed exceed launch speed?

The cliff added fall distance below the launch point (

Δ y < 0

), so

2 a Δ y

is positive and adds to

v_{0}^{2}

When do you keep both roots of the position quadratic?

When both are positive times and the object physically passes that height twice (target below the apex).

v_{0} \to 0

, what does max height

H = v_{0}^{2} /2 g

approach, and why does that make sense?

H \to 0

; a zero-speed "throw" is just a drop, which rises nothing.

Does the final answer depend on choosing up-positive vs down-positive?

No — Ex 4 and Ex 5 give identical depth and speed; only intermediate signs flip.

Connections

Parent: Free fall & sign conventions
Equations of Motion (constant acceleration) — the three tools used on every example
Vectors and sign conventions — why Ex 5 works
Projectile Motion — these vertical cases become the $y$ -component
Newton's Second Law — origin of $a = g$
Air Resistance and Terminal Velocity — when these idealised answers break

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