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

WHAT is free fall?

WHY mass doesn't matter: Newton says $F = ma$ . Gravity pulls with $F = mg$ . So $ma = mg \implies a = g.$ The mass cancels. Heavier objects feel more force, but also need more force to accelerate — these exactly offset.

HOW to set up sign conventions

Convention A — "Up is positive" (most common):

Acceleration $a = -g = -9.8\ \text{m/s}^2$ (gravity points down = negative).
Upward velocity is $+$ , downward velocity is $-$ .
Height above start is $+$ .

Convention B — "Down is positive" (handy for things only falling):

Acceleration $a = +g = +9.8\ \text{m/s}^2$ .
Downward velocity is $+$ .

Both give the same physics; pick whichever makes the initial conditions simplest.

Deriving the equations of motion from scratch

We start from constant acceleration $a$ (here $a = \pm g$ ). Acceleration is the rate of change of velocity: $a = \frac{dv}{dt}.$

Step 1 — velocity. Why? Integrate $dv = a\,dt$ with $a$ constant: $\int_{v_0}^{v} dv = \int_0^t a\,dt \implies v - v_0 = at \implies \boxed{v = v_0 + at}.$

Step 2 — position. Why? Velocity is $\dfrac{dy}{dt} = v_0 + at$ . Integrate: $\int_{y_0}^{y} dy = \int_0^t (v_0 + at)\,dt \implies y - y_0 = v_0 t + \tfrac12 a t^2.$ $\boxed{y = y_0 + v_0 t + \tfrac12 a t^2}.$

Step 3 — time-free relation. Why? Eliminate $t$ to relate $v$ and displacement. From Step 1, $t = (v-v_0)/a$ . Substitute into Step 2 and simplify: $\boxed{v^2 = v_0^2 + 2a\,\Delta y}.$

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

Key consequences (the 80/20 that solves most problems)

Time to top: set $v=0$ in $v = v_0 - gt \Rightarrow t_{\text{up}} = v_0/g$ .
Max height: $H = v_0^2/(2g)$ from $0 = v_0^2 - 2gH$ .
Up–down symmetry: time up = time down; the ball returns to launch height with speed $v_0$ (but downward, i.e. $-v_0$ ).

Worked examples

Common mistakes (steel-manned)

Recall Feynman: explain to a 12-year-old

Imagine you throw a ball straight up. Gravity is like an invisible hand that always pulls down, the same amount the whole time. On the way up the pull slows the ball; for one tiny instant at the top it's not moving; then the same pull speeds it up coming down. Every object — a bowling ball or a marble — gets sped up by the same amount each second (about 10 m/s faster every second), as long as there's no air pushing back. The "+ and –" we use is just a way of writing down "up" and "down" so the math doesn't get confused.

Active recall

What is free fall?

Motion under gravity alone (no air resistance/other forces); acceleration

=g\approx9.8\ \text{m/s}^2

downward.

Why do all masses fall with the same acceleration?

Because

ma=mg

⟹

a=g

; mass cancels.

With "up positive", what is the acceleration of a freely falling object?

a=-g=-9.8\ \text{m/s}^2

At the highest point of a vertical throw, what are velocity and acceleration?

v=0

, but

a=-g

(gravity still acts).

Time to reach the top when launched up at

v_0

t_{\text{up}}=v_0/g

Maximum height for launch speed

v_0

straight up?

H=v_0^2/(2g)

Three constant-acceleration equations (up positive)?

v=v_0-gt

;

y=y_0+v_0t-\tfrac12gt^2

;

v^2=v_0^2-2g\,\Delta y

A ball thrown up returns to launch height with what speed?

Same magnitude

v_0

but directed downward (

-v_0

Why isn't acceleration zero at the top of the path?

Velocity is momentarily zero but gravity never stops;

a=-g

continuously.

Connections

Equations of Motion (constant acceleration)
Vectors and sign conventions
Projectile Motion — free fall is the vertical component
Newton's Second Law — source of $a=g$
Air Resistance and Terminal Velocity — what breaks free fall

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Free fall ka matlab hai object sirf gravity ke under move kar raha hai — na air resistance, na koi push. Earth ke paas har object ka downward acceleration same hota hai, magnitude $g \approx 9.8\ \text{m/s}^2$ . Mass koi matter nahi karti, kyunki $ma = mg$ mein mass cancel ho jaati hai. Bhaari ho ya halka, vacuum mein dono ek saath girte hain — feather aur hammer same time mein!

Sabse important cheez hai sign convention. Pehle decide karo kaunsi direction positive hai. Agar "up = positive" lo, to gravity hamesha minus mein aati hai: $a = -9.8$ . Upar jaati velocity $+$ , neeche aati velocity $-$ . Agar object sirf gir raha hai to "down = positive" lena easy padta hai, tab $a = +9.8$ aur sab quantities positive — kam galti hoti hai. Bas ek convention pakdo aur poore problem mein wahi use karo, beech mein mat badlo.

Ek bada confusion: top point pe ball ruk jaati hai ( $v=0$ ), to log sochte hain $a$ bhi zero ho gaya. Galat! Gravity kabhi off nahi hoti, $a = -g$ chalti rehti hai — isiliye ball wapas neeche aati hai. Aur symmetry yaad rakho: upar jaane ka time = neeche aane ka time, aur ball wapas launch point pe same speed $v_0$ se aati hai (bas neeche direction mein).

Teen formulas se 80% problems ho jaate hain: $v=v_0-gt$ , $y=y_0+v_0t-\tfrac12 gt^2$ , aur $v^2=v_0^2-2g\Delta y$ . Time-free wala (teesra) tab use karo jab time nahi diya ho. Max height $H=v_0^2/2g$ aur top tak ka time $t=v_0/g$ — ye derive karna seekh lo, exam mein direct kaam aayega.

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

WHAT is free fall?

HOW to set up sign conventions

Deriving the equations of motion from scratch

Key consequences (the 80/20 that solves most problems)

Worked examples

Common mistakes (steel-manned)

Active recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Measurement, Vectors & Kinematics

Connections