Physics I: Mechanics › Kinematics › full formula sheet

y = v₀t − ½gt²Say it: height equals initial velocity times time, minus one half g t squared — with up positive and g pulling down

Free fall

The constant-acceleration equations with one substitution — a = −g — and the sign conventions that make or break every answer.

g = 9.8 m/s² downward, near Earth’s surface. With up positive: ay = −g, v₀ > 0 means thrown upward, y < 0 means below the launch point. Air resistance is neglected.

Where it comes from

Drop a stone down a well and time the splash: you can measure the well’s depth without a rope. Throw a ball straight up: you can predict exactly when it comes back. Both are free fall — motion under gravity alone — and both obey one equation. The surprise for beginners: “falling” includes going up.

Before reading on: you drop a ball from rest. After 1 s it has fallen 4.9 m. After 2 s — 9.8 m? Or something else? Guess, then watch the t² do its work.
t = 1 s
→
y = −½·9.8·1² = −4.9 m
Fallen 4.9 m below the drop point.
t = 2 s
→
y = −½·9.8·2² = −19.6 m
Not 9.8 m — four times as far. Doubling time quadruples the fall: the t² signature of acceleration.

Galileo’s insight (leaning-tower legend aside, his inclined-plane experiments were the real work): all objects fall at the same rate when air is out of the picture. One value of g, one equation, every projectile’s vertical motion.

Derivation

Free fall is not a new law — it is the constant-acceleration equations wearing gravity’s uniform. Substitute ay = −g (up positive) into each one:

ay
=
−g = −9.8 m/s²
Step 1 — the only physics input. Near Earth’s surface, gravity accelerates everything downward at 9.8 m/s². Mass cancels out (that’s why it’s the same for all objects).
v
=
v₀ + ayt = v₀ − gt
Step 2 — velocity. v = v₀ + at with a → −g. Thrown up at 20 m/s, after 3 s: v = 20 − 29.4 = −9.4 m/s (coming down).
y
=
v₀t + ½ayt² = v₀t − ½gt²
Step 3 — position. The hero equation. Dropped (v₀ = 0): y = −½gt² — pure fall.
v²
=
v₀² + 2ayΔy = v₀² − 2gΔy
Step 4 — time-free. Same substitution in v² = v₀² + 2aΔx. Handy when time is unknown.

Down-positive alternative: some textbooks take downward as positive, giving y = v₀t + ½gt² with v₀ > 0 meaning thrown down. Both are correct — pick one per problem and never mix them mid-solution.

How to use it

The procedure, every time:

  1. Fix the sign convention first — up positive is standard here. Then: v₀ > 0 = thrown up, v₀ = 0 = dropped, v₀ < 0 = thrown down. Gravity is always −g.
  2. Translate “dropped / thrown” into v₀. Dropped = released from rest (v₀ = 0), not “v₀ = g.”
  3. Pick the equation missing the variable you don’t need. Time unknown and unasked? Use v² = v₀² − 2gΔy.
  4. Read y correctly. y is measured from the launch point, not from the ground — unless the launch was at ground level.

Key results at a glance

tup = v₀/gtime to the peak — set v = v₀ − gt = 0Say it: time up equals launch speed over g
ymax = v₀²/(2g)peak height — the “how high” formula (see the max-height lesson for the angled version)

Thrown up at 20 m/s: tup = 20/9.8 ≈ 2.04 s, ymax = 400/19.6 ≈ 20.4 m.

Common mistake: writing ay = +g with up positive (“gravity is 9.8”). Then a dropped ball’s equation reads y = +½gt² — it rises. The sign on g is the single most failed detail in kinematics.
Common mistake: reporting “11 m above the ground” for a ball thrown from a 2 m balcony whose Δy = 11 m. y is displacement from launch; height above ground is 2 + 11 = 13 m. Always ask: “measured from where?”

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the basic move: dropped from 45 m, time to land?

  1. List givens (up positive). v₀ = 0 (dropped), Δy = −45 m (ends below launch), a = −g.
  2. Use y = v₀t − ½gt². −45 = 0 − 4.9t².
  3. Solve. t² = 45/4.9 = 9.184 → t = √9.184 ≈ 3.03 s.
  4. Check. ½·9.8·(3.03)² = 4.9 · 9.184 = 45.0 m ✓
Common mistake: setting Δy = +45 (“the height is 45”). Then −4.9t² = +45 has no real solution — the equation is telling you the signs are inconsistent. Below launch = negative.
Your turn — dropped from 20 m. Time to hit the ground?

Answer: 2.02 s. t² = 20/4.9 = 4.0816; t = √4.0816 ≈ 2.02 s.

Example 2 — thrown up: v₀ = 20 m/s, time to peak and max height

  1. Time to peak: at the top v = 0, so 0 = 20 − 9.8t → t = 20/9.8 ≈ 2.04 s.
  2. Max height: y = v₀t − ½gt² = 20·2.0408 − 4.9·(2.0408)².
  3. Compute. 40.816 − 4.9·4.165 = 40.816 − 20.408 = 20.4 m.
  4. Cross-check with ymax = v₀²/(2g). 400/19.6 = 20.4 m ✓
Common mistake: computing y at t = 2.04 s but with +½gt² (wrong sign on g): 40.8 + 20.4 = 61.2 m — higher than the launch speed could ever reach. The minus is load-bearing.
Your turn — thrown up at 14.7 m/s. Time to peak? Max height?

Answer: 1.5 s; 11.0 m. t = 14.7/9.8 = 1.5 s; y = 14.7 · 1.5 − 4.9 · 2.25 = 22.05 − 11.025 = 11.025 m.

Example 3 — total hang time: thrown up at 20 m/s

  1. Use symmetry. With no air drag, the trip up mirrors the trip down: ttotal = 2 · tup.
  2. From Example 2: tup ≈ 2.04 s.
  3. Double. ttotal = 4.08 s to return to the launch height.
  4. Verify with the equation. Set y = 0: 0 = 20t − 4.9t² = t(20 − 4.9t) → t = 0 (launch) or t = 20/4.9 = 4.08 s ✓.
Common mistake: solving y = 0 and keeping only t = 0 (“it never lands”). The factored form t(20 − 4.9t) = 0 has two roots — launch and landing. Always check for the second root.
Your turn — thrown up at 9.8 m/s. Total hang time back at launch height?

Answer: 2 s. tup = 9.8/9.8 = 1 s; double → 2 s. (Or 0 = 9.8t − 4.9t² → t = 2 s.)

Before reading on: throwing the ball downward at 5 m/s from 30 m up — will it land in more or less than the 2.47 s a dropped ball takes? (Dropped: t = √(2·30/9.8) ≈ 2.47 s.) Estimate, then compute.

Example 4 — thrown down: v₀ = −5 m/s from 30 m up

  1. List givens (up positive). v₀ = −5 m/s (thrown down), Δy = −30 m.
  2. Equation: −30 = −5t − 4.9t² → 4.9t² + 5t − 30 = 0.
  3. Quadratic formula. t = [−5 + √(25 + 588)]/9.8 = (−5 + √613)/9.8 = (−5 + 24.76)/9.8 ≈ 2.02 s. (The other root is negative — reject it.)
  4. Sense-check. 2.02 s < 2.47 s (dropped) ✓ — the head start helped, as predicted.
Common mistake: writing v₀ = +5 (“the speed is 5”). Thrown down with up positive means v₀ = −5. Speed is 5 m/s; velocity is −5 m/s.
Your turn — thrown down at 3 m/s from 20 m up. Time to land?

Answer: 1.74 s. 4.9t² + 3t − 20 = 0; t = (−3 + √401)/9.8 = (−3 + 20.02)/9.8 ≈ 1.74 s.

Memorization tips

  • Say it aloud: “y equals v-naught t minus one-half g t squared.” The minus is the whole battle — say it every time.
  • Sign convention first, numbers second. Write “up = +” at the top of every free-fall problem. Half of all errors die here.
  • Dropped = v₀ = 0. Not v₀ = g, not v₀ = “small” — zero. Thrown = nonzero v₀ with the throw’s sign.
  • tup = v₀/g and ymax = v₀²/(2g) — the two shortcuts for “thrown straight up” problems. Derive them once, then use freely.
  • Symmetry is free: no air drag → up-time = down-time, and it returns at the same speed (opposite direction). Use it as a check.
  • g = 9.8, not 10 — unless the problem says otherwise. The 2% matters on exams that check exact answers.

Final challenge

Five mixed questions — basics, applications, and the traps, all in one. Score 5/5 and Free fall is yours.

← Back to the Physics I formula sheet

How to learn a formula here

  1. Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
  2. Work the examples with the answers covered, then uncover one step at a time and compare.
  3. Finish with the final challenge — five mixed questions including the classic traps.
  4. Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

What is free fall?

Motion under gravity alone, with no other forces (air resistance neglected). Near Earth's surface that means constant downward acceleration g = 9.8 m/s^2 — the same for every object, which is why a dropped stone and a thrown ball share one equation.

Is going up considered free fall?

Yes. 'Free fall' means gravity is the only force, not 'moving downward'. A ball tossed upward is in free fall the whole way up and down — with a_y = -g throughout, even at the peak where v = 0.

Why is g negative in the equations?

With the standard 'up positive' convention, gravity points down, so its acceleration is -9.8 m/s^2. The negative sign is what makes dropped balls fall and tossed balls come back. (Taking down as positive flips all the signs — also fine, but pick one.)

Do heavier objects fall faster?

No — not in free fall. g is independent of mass (Galileo's insight, confirmed by the famous hammer-feather drop). In air, drag affects light/flat objects more, which is why real feathers flutter — that's air resistance, not gravity.

How long does it take to fall a height h?

From rest: h = 1/2 g t^2, so t = sqrt(2h/g). From 45 m that's about 3.03 s. Note the square root: falling 4x as far takes only 2x as long.

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