Free Fall & Motion Under Gravity

Drop your phone and your textbook from the same height (please don't) and they hit the floor together. That flat fact took two thousand years to establish, cost Aristotle his authority, and eventually rebuilt our entire picture of gravity. Aristotle's claim — heavy things fall faster, in proportion to weight — sounds so reasonable that nobody bothered to test it until Galileo demolished it twice: once with an argument so clean it needs no apparatus at all, and once with bronze balls rolled down ramps, timed against a water clock. What he found is the founding measurement of modern physics: near the Earth's surface, every body falls with the same constant acceleration, $g$ . For your exams, this is the best news in kinematics. Free fall introduces no new mathematics — it is the suvat toolkit you already own, with the acceleration filled in for you. The entire skill is bookkeeping: pick a sign convention, keep it, and let the up–down symmetry of the flight cut your work in half. And hiding inside the exam topic is one of the deepest ideas in physics. Because everything falls alike, a freely falling observer feels no gravity at all — the thought Einstein later called the happiest of his life, and the reason astronauts float. Master this module and you hold both: the marks, and the seed of general relativity.

Key takeaways

Free fall = motion under gravity alone (no drag, no thrust, no contact forces). Near Earth's surface it is constant-acceleration motion with $a = g \approx 9.8\ \mathrm{m/s^2}$ directed downward — for every object, whatever its mass.
All of suvat applies unchanged: with up as positive, $v = u - gt$ , $s = ut - \tfrac12 gt^2$ , $v^2 = u^2 - 2gs$ . Gravity hands you the acceleration; you only need two more quantities.
Up–down symmetry (no air resistance): time up = time down, and the object returns to any height with the same speed it had there on the way up. One symmetry line often replaces half a page of algebra.
At the highest point $v = 0$ but $a$ is still $g$ downward — the acceleration never pauses, which is exactly why the ball comes back.
From rest, distances fallen in successive seconds run $1:3:5:7$ (Galileo's odd-number rule); with $g = 10$ that is $5, 15, 25, 35$ m, totals $5, 20, 45, 80$ m.
Exam convention: CBSE, JEE and NEET numericals usually take $g = 10\ \mathrm{m/s^2}$ unless stated; AP, IB, IGCSE and A-Level papers work with $9.8$ . Check before you compute.
Key idea: gravity pulls in proportion to exactly the mass that inertia divides by, so the mass cancels — free fall is a property of the place, not of the object.

The core insight

In free fall the only force is gravity, and because the mass being pulled equals the mass resisting acceleration, every body accelerates at the same $g \approx 9.8\ \mathrm{m/s^2}$ downward throughout the motion — while rising, at the momentary stop at the top, and while falling — so one constant-acceleration setup with a fixed sign convention solves every problem in the chapter.

Where this lives today

Free fall is now a manufactured laboratory condition. The ZARM drop tower in Bremen lets sealed capsules fall through a 120-metre evacuated shaft, buying about 4.7 seconds of clean weightlessness for experiments in combustion, fluids and cold atoms; parabolic "vomit comet" flights buy ~20 seconds; the International Space Station supplies the condition indefinitely — which is precisely why it exists as a laboratory. "Microgravity" means Galileo's condition: everything falling together, so nothing presses on anything. The demonstration got its definitive television remake in 2014, when Brian Cox dropped a bowling ball and feathers inside NASA's giant vacuum chamber in Ohio — the feathers fell like stones. The frontier has swapped bronze balls for atoms: interferometers drop clouds of two rubidium isotopes in one vacuum tube and compare their accelerations by matter-wave interference, testing universality at the $10^{-12}$ level in quantum systems. Any confirmed violation, at any decimal place, would be the first crack in general relativity — which is why the oldest experiment in physics keeps being rebuilt with ever-better feathers.

The physical picture

Throw a ball straight up at $20\ \mathrm{m/s}$ and watch what gravity does to the velocity, second by second (take $g = 10\ \mathrm{m/s^2}$ for clean numbers). Launch: $20\ \mathrm{m/s}$ up. One second later: $10$ up. At $t = 2\ \mathrm{s}$ : zero — the ball hangs, for one instant, at its highest point, $20\ \mathrm{m}$ up. Then the movie runs in reverse: $10\ \mathrm{m/s}$ down at $t = 3\ \mathrm{s}$ , $20$ down at $t = 4\ \mathrm{s}$ as it returns to your hand. Every second, rising or falling, the velocity changes by the same $10\ \mathrm{m/s}$ downward. That is the whole physics: one constant downward acceleration, acting identically through the entire flight.

Figure

One throw, one acceleration — the up-down symmetric flight of a ball

Two features of this picture do most of the exam work. First, the symmetry: the flight down is the flight up played backwards — time up equals time down, and the speed at any height going down equals the speed there going up. Second, the apex trap: at the top the velocity is zero but the acceleration is not; if acceleration vanished there, nothing would change the zero velocity and the ball would hang forever. It comes back precisely because $g$ keeps acting. But why do a cricket ball and a marble — different masses, different materials — trace exactly the same flight? Galileo answered with pure logic before anyone could answer with instruments.

Derivation AGalileo's tied-stones argument — heavier cannot fall faster

We want to show that the fall rate cannot depend on weight, and we will do it Galileo's way: assume the opposite and watch the assumption destroy itself — a proof by contradiction needing no apparatus at all. Watch how one physical setup gets two incompatible predictions.

Step 1. Assume Aristotle is right: a heavy stone falls faster than a light one, in proportion to weight. Tie a heavy stone H and a light stone L together with a short cord.

Step 2. Predict from the parts: L, falling slower, lags and pulls back on H through the cord — the pair should fall slower than H alone.

Step 3. Predict as a whole: tied together, H and L are one body heavier than H — the pair should fall faster than H alone.

Step 4. One experiment, two contradictory predictions. The assumption dies. The only escape is that all bodies fall at the same rate — and the cord then carries no tension at all. Verified: the modern statement is that gravitational mass (what gravity pulls) equals inertial mass (what resists acceleration), so the mass cancels in the second law and a = g for every body — tested by the MICROSCOPE satellite in 2022 to one part in 10¹⁵.

What everyday experience actually shows is air resistance. A feather in air is a sail with almost no weight behind it; it loses to drag, not to weaker gravity. Remove the air and the race is always a tie: in 1971 astronaut David Scott, on the airless Moon, released a hammer and a falcon feather together on live television — they struck the dust in the same instant. In exam language: falling with air resistance is qualitative territory (the object speeds up, drag grows with speed, acceleration shrinks, and the fall settles at a constant terminal velocity); falling without it — "free fall" — is where all the calculations live.

The math: suvat with the acceleration filled in

Free fall is strictly one-dimensional motion, so the first decision of every problem is a sign convention. Take upward as positive — the most common choice — and the acceleration becomes $a = -g$ for the whole flight. The three constant-acceleration equations you already own then read:

v = u - gt \qquad (1)

where: $u$ — launch velocity (positive up); $v$ — velocity at time $t$ ; $g$ — the magnitude $9.8\ \mathrm{m/s^2}$ (or $10$ in CBSE/JEE/NEET numericals). Equation (1) says velocity drops by $g$ every second: a straight line on the v–t graph, crossing zero at the apex. Nothing about the object's mass appears anywhere — that is Galileo's discovery written as algebra.

s = ut - \tfrac{1}{2} g t^2 \qquad (2)

where: $s$ — the displacement from the launch point, positive if above it. This is Galileo's square law: for a pure drop (down positive, $u = 0$ ) it collapses to the chapter's best mental formula, $h = \tfrac12 gt^2 \approx 5t^2$ metres with $g = 10$ . Three seconds of falling means $45$ m — an estimate you can do while reading the question.

v^2 = u^2 - 2gs \qquad (3)

where: $v$ — velocity at displacement $s$ . The clockless equation: it links speed to height directly and is energy conservation in disguise. Set $v = 0$ and it hands you the maximum height $H = u^2/2g$ ; run it from the apex downward and it returns the landing speed. The other convention — downward positive, natural for drops — just flips every sign: $v = u + gt$ , $s = ut + \tfrac12 gt^2$ , $v^2 = u^2 + 2gs$ . Both give identical answers; every sign error in this chapter comes from mixing them mid-problem, not from choosing the "wrong" one.

Figure

Two sign conventions, one answer — a 45 m drop solved both ways

Which board uses which $g$ ? CBSE, JEE-Main/Advanced and NEET numericals take $g = 10\ \mathrm{m/s^2}$ unless the paper says otherwise; AP, IB, IGCSE and A-Level use $9.8$ ( $9.81$ in some A-Level data booklets). State your value once at the top of a written answer and no examiner can fault you.

Go deeper · Derivation B · 1 minUp-down symmetry — time up equals time down, return speed equals launch speed

Derivation BUp-down symmetry — time up equals time down, return speed equals launch speed

We want the two symmetry results that let you solve half of all free-fall problems in one line: for a launch at speed u (no air resistance), the time to fall from the apex back to the launch height equals the time to rise, and the return speed equals u. The strategy is to compute the up leg and the down leg separately with the suvat set and watch them come out identical. The thing to watch: we use the same g for both legs — this is exactly where air resistance would break the symmetry.

Step 1. The rise. From equation (1), the apex ( $v = 0$ ) arrives at $t_{\text{up}} = u/g$ . From equation (3), the apex height is $H = u^2/2g$ .

Step 2. The fall. Restart the clock at the apex: the object falls from rest through the same height H. Equation (2) with down positive gives $H = \tfrac12 g\, t_{\text{down}}^2$ , so

t_{\text{down}} = \sqrt{\frac{2H}{g}} = \sqrt{\frac{2}{g}\cdot\frac{u^2}{2g}} = \frac{u}{g}

Why it matched: H itself was built from u and g by the deceleration; feeding it back into the same-g fall undoes the algebra exactly. Symmetry is not an axiom here — it is a consequence of ONE unchanging acceleration.

Step 3. The return speed. Falling from rest through H: $v^2 = 2gH = 2g \cdot u^2/2g = u^2$ , so the landing speed is exactly $u$ , directed downward.

Step 4. Collect the flight facts for a throw from ground level: $t_{\text{up}} = t_{\text{down}} = u/g$ , total time of flight $T = 2u/g$ , maximum height $H = u^2/2g$ , return speed $= u$ . Verified: u = 20 m/s, g = 10 gives t_up = 2 s, H = 20 m, T = 4 s, return speed 20 m/s — matching the second-by-second table in the physical picture above.

Go deeper · Derivation C · 1 minThe odd-number rule — why successive seconds cover distances 1 : 3 : 5 : 7

Derivation CThe odd-number rule — why successive seconds cover distances 1 : 3 : 5 : 7

We want Galileo's signature pattern: from rest, the distances fallen in the 1st, 2nd, 3rd, ... seconds are in the ratio 1 : 3 : 5 : 7. This is worth deriving because examiners test it directly ("distance in the nth second") and because it is the fastest sanity check on any table of fall data. The strategy: subtract total distances at successive whole seconds — pure algebra on Galileo's square law.

Step 1. From rest (down positive), the total distance by time t is $h(t) = \tfrac12 g t^2$ . At $t = 1, 2, 3, 4$ s the totals are $\tfrac12 g \times 1, 4, 9, 16$ — perfect squares, the fingerprint of constant acceleration.

Step 2. The distance covered during the nth second alone is the difference of totals:

d_n = \tfrac{1}{2} g\, n^2 - \tfrac{1}{2} g\, (n-1)^2 = \tfrac{1}{2} g\,(2n - 1)

Why the odd numbers appear: the difference of consecutive squares, n² − (n−1)², is always the odd number 2n − 1. The physics (constant g) enters only through the square law; the odd numbers are arithmetic.

Step 3. Read the ratio: $d_1 : d_2 : d_3 : d_4 = 1 : 3 : 5 : 7$ . With $g = 10$ : $5, 15, 25, 35$ m in successive seconds, totals $5, 20, 45, 80$ m. Verified: sum of the first four odd multiples 5 + 15 + 25 + 35 = 80 m equals the direct total ½ × 10 × 4² = 80 m. Galileo read exactly this spacing off his inclined ramp in 1604 — the ramp only dilutes g, leaving the ratio untouched.

Limiting cases

case	result	meaning
$u = 0$ (pure drop)	$h = \tfrac12 gt^2 \approx 5t^2$ m, $v = gt$	the mental-math workhorse: 5, 20, 45, 80 m
at the apex	$v = 0$ , $a = g$ downward	velocity pauses; acceleration never does
return to launch height	speed $= u$ , time $= 2u/g$	perfect up–down symmetry (no air)
strong air resistance	$a \to 0$ , $v \to$ terminal velocity	drag grows with speed until it balances weight
very large heights	$g$ itself shrinks as $GM/(R+h)^2$	"constant g" is the local, near-surface approximation
falling observer	apparent weight zero	free fall feels like no gravity — Einstein's seed

Common trap“At the highest point velocity is zero, so acceleration is zero”

The single most-tested misconception in this chapter, and assertion–reason papers are built on it. At the apex the ball's velocity passes through zero, but gravity has not been switched off: the acceleration is $g \approx 9.8\ \mathrm{m/s^2}$ downward at that instant, exactly as it is everywhere else in the flight. The proof is one thought: if $a$ were zero when $v$ is zero, nothing would ever change the zero velocity, and the ball would hover forever. It comes down because the acceleration persists. Keep the two ideas separate: velocity says how position is changing; acceleration says how velocity is changing — and a quantity can be zero at an instant while changing at full rate. On the v–t graph the line crosses the axis at the apex without even flinching: same slope $-g$ before, at, and after the crossing.

Worked examples

Board level — stone thrown up from a tower (CBSE/ICSE pattern)

A stone is thrown vertically upward at $20\ \mathrm{m/s}$ from the top of a tower $25\ \mathrm{m}$ high. Find (a) the maximum height above the ground, (b) the total time before it hits the ground, and (c) its speed on impact. Take $g = 10\ \mathrm{m/s^2}$ .

Step 1. Set the convention: up positive, origin at the top of the tower. Then $u = +20\ \mathrm{m/s}$ , $a = -10\ \mathrm{m/s^2}$ , and the ground is at $s = -25\ \mathrm{m}$ .

Step 2. Maximum height, equation (3) with $v = 0$ : $0 = 20^2 - 2(10)H \Rightarrow H = 400/20 = 20\ \mathrm{m}$ above the tower top, i.e. $20 + 25 = \boxed{45\ \mathrm{m}}$ above the ground.

Step 3. Total time, equation (2) with $s = -25$ : $-25 = 20t - 5t^2$ , i.e. $t^2 - 4t - 5 = 0 = (t-5)(t+1)$ . The physical root is $\boxed{t = 5\ \mathrm{s}}$ ; discard $t = -1$ s with one written word ("non-physical").

Step 4. Impact speed, equation (1): $v = 20 - 10(5) = -30\ \mathrm{m/s}$ — magnitude $\boxed{30\ \mathrm{m/s}}$ , the sign confirming it points downward. Cross-check with (3) from the apex: falling from rest through 45 m, $v = \sqrt{2 \times 10 \times 45} = 30\ \mathrm{m/s}$ ✓.

Step 5. With $g = 9.8$ (AP/IB/A-Level flavour): $4.9t^2 - 20t - 25 = 0$ gives $t = (20 + \sqrt{890})/9.8 = 5.08\ \mathrm{s}$ , impact speed $\sqrt{890} = 29.8\ \mathrm{m/s}$ — same method, heavier arithmetic.

Competitive level — the last-second fraction (NEET/JEE-Main pattern)

A body falls freely from rest and covers half of its total height in the last second of its fall. Find the total time of fall and the height. ( $g = 10\ \mathrm{m/s^2}$ )

Step 1. Name the total time $T$ . Total height: $h = 5T^2$ . Distance in the last (i.e. the $T$ th) second, from the odd-number result: $d_T = 5(2T - 1)$ .

Step 2. Impose $d_T = h/2$ : $5(2T - 1) = \tfrac12 \times 5T^2 \Rightarrow 2(2T-1) = T^2 \Rightarrow T^2 - 4T + 2 = 0$ .

Step 3. Solve: $T = \dfrac{4 \pm \sqrt{16 - 8}}{2} = 2 \pm \sqrt{2}$ . Both roots are positive — now think. The condition "covers half the height in the last second" needs the fall to last at least one second, and the root $2 - \sqrt{2} = 0.59\ \mathrm{s} < 1\ \mathrm{s}$ makes "the last second" longer than the whole fall. Reject it. $\boxed{T = 2 + \sqrt{2} \approx 3.41\ \mathrm{s}}$ .

Step 4. Height: $h = 5T^2 = 5(6 + 4\sqrt2) = 30 + 20\sqrt{2} \approx \boxed{58.3\ \mathrm{m}}$ . Check: last-second distance $5(2 \times 3.414 - 1) = 29.1\ \mathrm{m}$ — half of $58.3$ ✓. Timing note: 90 seconds. The entire question is knowing $d_n = \tfrac12 g(2n-1)$ cold and rejecting the spurious root with a one-line reason. If you derive the nth-second formula from scratch in the exam, you have donated two minutes.

Competitive level — passing the same height twice (JEE-Main pattern)

A ball thrown vertically upward passes a point P at height $h$ at times $t_1 = 2\ \mathrm{s}$ and $t_2 = 6\ \mathrm{s}$ after launch. Find the launch speed, the height of P, and the maximum height. ( $g = 10\ \mathrm{m/s^2}$ )

Step 1. Both crossings satisfy equation (2): $h = ut - 5t^2$ , i.e. $5t^2 - ut + h = 0$ . The two times $t_1, t_2$ are the two roots of the same quadratic — that observation is the whole trick.

Step 2. Use the root relations. Sum: $t_1 + t_2 = u/5 \cdot 1 \Rightarrow u = 5(t_1 + t_2)$ ... write it cleanly: sum of roots $= u/5$ , so $u = 5(2 + 6) = \boxed{40\ \mathrm{m/s}}$ . Product of roots $= h/5$ , so $h = 5\,t_1 t_2 = 5 \times 12 = \boxed{60\ \mathrm{m}}$ .

Step 3. Maximum height: $H = u^2/2g = 1600/20 = \boxed{80\ \mathrm{m}}$ . Sanity checks: by symmetry the apex is at $t = (t_1 + t_2)/2 = 4\ \mathrm{s} = u/g$ ✓; direct substitution $s(2) = 80 - 20 = 60$ m ✓ and $s(6) = 240 - 180 = 60$ m ✓. Timing note: 60–75 seconds via sum-and-product of roots; 4+ minutes if you solve the two suvat equations simultaneously. The general results $u = \tfrac12 g(t_1{+}t_2)$ and $h = \tfrac12 g\,t_1 t_2$ are worth memorising — they recur yearly in JEE-Main and NEET banks.

Exam traps

Flipping the sign of g mid-problem. The killer. Choose up-positive (then $a = -g$ for the whole flight) or down-positive (then $a = +g$ ), write it at the top, and never renegotiate — not at the apex, not when the ball starts descending. The acceleration does not change direction when the velocity does.

"At the top, a = 0." Assertion–reason and MCQ papers plant this every year. At the apex $v = 0$ momentarily but $a = g$ throughout. If the option says the acceleration at the highest point is zero, it is wrong — full stop.

Using the wrong g for your board. CBSE/JEE/NEET build options from $g = 10$ ; AP, IB, IGCSE and A-Level mark schemes expect $9.8$ (A-Level data booklets often say $9.81$ ). With the wrong constant your answer lands between options and you burn a minute doubting correct work.

Displacement is not distance on a round trip. Thrown up 20 m and caught back: displacement $0$ , distance $40\ \mathrm{m}$ . The suvat equations use displacement — for total distance, split the journey at the apex.

Treating the throw as two separate problems by default. Symmetry usually does it in one line: $t_{\text{up}} = u/g$ , return speed $= u$ . Splitting doubles the algebra and the sign-error surface — split only when the legs genuinely differ (thrown from a tower, landing below launch height).

Time to fall h is not h/g. It is $t = \sqrt{2h/g}$ — the square root catches students who pattern-match from $v = u + at$ . Quick anchor: 45 m takes 3 s, not 4.5 s.

Forgetting the air-resistance caveat. Symmetry results hold only for negligible drag. With air resistance, time down exceeds time up and return speed < launch speed — IGCSE and AP love this qualitative twist.

Origins · 1 minAlbert Einstein and how this was discovered

Albert Einstein

1879–1955 · Bern, Zurich, Princeton

In 1907, still at the Bern patent office, Einstein had what he later called "the happiest thought of my life": a person falling freely does not feel their own weight. Drop the observer along with the laboratory and gravity locally disappears — keys float, a bathroom scale reads zero. That could only be true because everything falls at exactly the same rate; the tiniest mass-dependence in $g$ and the falling observer would see objects drift apart. Einstein promoted Galileo's old measurement to a principle — the equivalence principle — and eight years later it had become general relativity, where gravity is the shape of spacetime and a freely falling ball moves as straight as the universe allows. Every astronaut floating in the ISS is a demonstration: orbit is not the absence of gravity, it is permanent free fall.

History

1589

— Legend places Galileo atop the Leaning Tower of Pisa, dropping unequal masses that land together; Simon Stevin had performed the Delft church-tower version in 1586.

1604

— Galileo extracts the times-squared law and the odd-number spacing 1 : 3 : 5 from inclined-plane experiments — gravity diluted on a ramp, not altered.

1638

— Two New Sciences publishes the ramp data and the tied-stones argument, retiring Aristotle's proportional-to-weight fall.

1907

— Einstein's "happiest thought": free fall locally cancels gravity — the equivalence principle that grows into general relativity.

1971

— Apollo 15: David Scott drops a hammer and a falcon feather on the Moon; they hit the dust together on live television.

2022

— The MICROSCOPE satellite's final results confirm that titanium and platinum fall alike to one part in 10¹⁵.

Going further

Texts. NCERT Physics XI Part 1, Ch. 3 (Motion in a Straight Line, free-fall section) — the CBSE/NEET canon · H.C. Verma, Concepts of Physics Vol. 1, Ch. 3 · Galileo, Two New Sciences, Third Day — the odd-number rule and the tied-stones argument in the master's own dialogue · Halliday, Resnick & Walker, Ch. 2. The UG deep dive is Free Fall and the Equivalence Principle territory: see the mechanics tier's gravitation articles (aq-mech tier).

Provenance. Adapted from the approved v3 pilot on free fall (Galileo's insight arc, tied-stones logic, MICROSCOPE bound) with the HS exam apparatus added. A verifier should recompute: the flight table for $u = 20$ , $g = 10$ ( $t_{\text{up}} = 2$ s, $H = 20$ m, $T = 4$ s); odd-number distances $5, 15, 25, 35$ m, totals $5, 20, 45, 80$ m; the tower example ( $t = 5$ s, impact 30 m/s; with $g = 9.8$ : 5.08 s, 29.8 m/s); the last-second problem ( $T = 2{+}\sqrt2 = 3.41$ s, $h = 58.3$ m); the two-crossing problem ( $u = 40$ m/s, $h = 60$ m, $H = 80$ m); ISS $g \approx 8.7\ \mathrm{m/s^2}$ at 400 km.

Prerequisites. Kinematic Equations for Uniform Acceleration — free fall is those equations with $a$ filled in. Leads to. Two Bodies Under Gravity — the drop-and-throw meeting archetype · Projectile Motion — free fall plus a horizontal coast.

Does a heavier ball really not fall faster? Feathers clearly fall slower.

In air, light-and-fluffy loses to drag, not to weaker gravity. The feather presents a large area with almost no weight behind it, so air resistance balances its tiny weight at a very low speed and it drifts down at terminal velocity. Remove the air and the race is a tie every time — hammer and feather on the Moon (1971), bowling ball and feathers in NASA's vacuum chamber (2014). For dense, compact objects at everyday speeds (stones, cricket balls, your phone), drag is negligible for the first several metres and "free fall" is an excellent model — which is what exam questions assume unless they explicitly say otherwise.

Is g positive or negative? Different books seem to disagree.

$g$ itself is a magnitude — $9.8\ \mathrm{m/s^2}$ , always positive, always meaning "the size of the free-fall acceleration". The sign of the acceleration in your equations depends on the axis you chose: up-positive gives $a = -g$ ; down-positive gives $a = +g$ . The books don't disagree about physics, only about axis choice. Write your convention at the top of the solution, substitute once, and never touch the sign again — including at the apex, including when the motion reverses. Mixing conventions mid-problem is the most common self-inflicted wound in this chapter.

If velocity is zero at the top, why doesn't the ball just stay there?

Because acceleration is not zero there. Velocity tells you how position is changing; acceleration tells you how velocity is changing — and at the apex the velocity is zero but changing at the full rate of $9.8\ \mathrm{m/s}$ every second. One instant later the ball has a small downward velocity, and down it comes. If the acceleration were zero at the top, you would be right: nothing would ever change the zero velocity and the ball would hover forever. The fact that it returns is the proof that gravity never paused. On the v–t graph, the line crosses zero without changing slope.

My mock test used g = 9.8 but my coaching material uses 10. Which should I use?

Match the paper, not the habit. CBSE, JEE-Main/Advanced and NEET numericals are built with $g = 10$ unless the question states otherwise — their options come out as clean integers (45 m, 3 s, 30 m/s). AP, IB, IGCSE and A-Level expect $9.8$ (sometimes $9.81$ per the data booklet). A value supplied in the question overrides everything. If your computed answer sits between two options, suspect the constant before the physics.

Astronauts float in the ISS — is there no gravity up there?

There is plenty: at ISS altitude ( $\sim 400$ km), $g$ is still about $8.7\ \mathrm{m/s^2}$ — nearly 90% of its surface value. Astronauts float because the station and everything in it are in permanent free fall together, falling around the Earth while moving sideways fast enough to keep missing it. Nothing presses on anything, scales read zero — Einstein's equivalence principle in action. "Weightlessness" means "freely falling", not "gravity-free"; the same thing happens for one second in a dropping lift or over a rollercoaster crest.

Does time up really equal time down if there is air resistance?

No — the symmetry is a vacuum result. With drag, the ball rises against gravity plus drag (large deceleration, shorter time and height), but descends with gravity minus drag (smaller acceleration). It returns moving slower than launch, and the descent takes longer than the ascent. Exams use this as a qualitative twist: "state and explain how air resistance changes the time of flight". The safe two-liner: drag always opposes motion, so it steals mechanical energy on both legs — speeds are lower at every height on the way down, hence more time falling.

Why is the distance in the nth second not the distance in n seconds?

"In $n$ seconds" means the total from the start: $\tfrac12 g n^2$ . "In the nth second" means during that one-second window alone: total at $n$ minus total at $n{-}1$ , which is $\tfrac12 g (2n - 1)$ — the odd-number formula. With $g = 10$ : totals go 5, 20, 45, 80 m while individual seconds go 5, 15, 25, 35 m. Misreading this phrase is a guaranteed lost mark, and setters know it: both numbers usually appear among the options.