ARCHETYPE: Two Bodies Under Gravity

Legend says Galileo dropped two balls — one heavy, one light — from the Leaning Tower of Pisa and watched them land together. Whether or not he ever climbed that tower, the idea he proved is the engine of an entire family of exam problems: gravity accelerates every free body identically. Drop a stone and throw another straight up past it; juggle two balls a second apart; release a packet from a rising balloon — in every case, each body obeys the same $-g$ , no matter which way it happens to be moving. That shared acceleration is a gift examiners cannot resist. A ball dropped from a tower and a ball thrown up from the ground must meet — but when, and where? Two balls thrown up with the same speed one second apart cross at a height that looks magical until you see the symmetry. A packet "dropped" from a climbing balloon first rises — the single most reliable trap in kinematics. CBSE and ISC pose the tower meeting as a 3-mark structured question; JEE-Main and NEET compress it into 90-second MCQs; JEE-Advanced stacks conditions ("they meet exactly when the second ball peaks…") that fall apart instantly once you know the master move. And the master move is almost embarrassingly simple: because both bodies share the same $-g$ , their relative acceleration is zero — so the gap between them closes at constant speed. This article builds that machine: one axis, one clock, one sign convention, position functions for each body, and the relative shortcut that turns four-line solutions into one.

Key takeaways

Two bodies moving vertically under gravity share the same acceleration $-g$ — regardless of their velocities' directions.
Master setup: ONE origin, ONE positive direction (up), ONE clock. Each body gets a position function $y_i(t) = y_{0i} + u_i t - \tfrac{1}{2}g t^2$ ; they meet when $y_1 = y_2$ .
The killer shortcut: relative acceleration is zero, so the gap closes at constant relative speed — meeting time $=$ initial gap $\div$ relative speed.
Drop from height $H$ + simultaneous upward throw at $u$ : they meet at $t = H/u$ , always.
A body released from a rising platform inherits the platform's upward velocity — it rises before it falls.
Key idea: shared $-g$ makes the relative motion of two free bodies uniform — set the position functions equal, or let the gap close at constant speed.

The core insight

Because every freely moving body near Earth has the same acceleration $-g$ , two such bodies have zero relative acceleration: write each position as $y_i(t) = y_{0i} + u_i t - \tfrac{1}{2}gt^2$ on one shared axis and clock, and their separation changes at the constant rate $u_1 - u_2$ .

Where this lives today

Equal fall under shared gravity is now a precision industry. Drop towers like ZARM in Bremen give experiments seconds of free fall in an evacuated shaft — everything in the capsule, released together, floats together, exactly the zero-relative-acceleration theorem of Derivation C. The International Space Station is the same physics in permanent flight: every body aboard shares one acceleration, so astronauts float beside their spanners. And the MICROSCOPE satellite mission (2017–2022) confirmed that different materials fall alike to a few parts in $10^{15}$ . Einstein promoted Galileo's equal fall into the equivalence principle, the cornerstone of general relativity; your tower-meeting problem is the first rung of that ladder.

The setup

Geometry first, because in an archetype the geometry is the physics. Everything happens on one vertical line. Fix the origin at the ground, take up as positive, and start one clock at the first release. Every body in the problem — dropped, thrown up, thrown down, released from a balloon — is then a single function of the same $t$ :

y_i(t) = y_{0i} + u_i\,t - \tfrac{1}{2}\,g\,t^2 \qquad (1)

where: $y_{0i}$ — where body $i$ starts; $u_i$ — its initial velocity with sign (up $+$ , down $-$ , dropped $= 0$ , released from a platform $=$ the platform's velocity); $g$ — magnitude of gravitational acceleration, entering with a minus because up is positive. Read it physically: the only thing distinguishing the bodies is their launch data $(y_{0i}, u_i)$ ; the $-\tfrac{1}{2}gt^2$ term is identical for all of them — that shared term is the entire secret of this archetype.

Figure

Tower meeting — dropped ball and thrown ball on one axis, one clock

Two disciplines make the machine run, and both earn explicit statements in a board answer:

One sign convention, declared once. Choose up as positive and write it down. Then $g$ enters every equation as $-9.8\ \mathrm{m/s^2}$ (or $-10$ ), a dropped body has $u = 0$ , a thrown-down body has negative $u$ — and you never patch signs mid-solution. Mixing conventions between the two bodies is the number-one mark-killer in this archetype.

One clock. Start $t = 0$ at the first event. A body launched later, at $t = t_0$ , has been flying for $(t - t_0)$ , so its position function uses $(t - t_0)$ everywhere $t$ would appear. Two private clocks — one per ball — is how juggler problems go wrong. The payoff for this discipline is the archetype's superpower. Subtract two copies of equation (1): the $-\tfrac{1}{2}gt^2$ terms cancel. The separation $y_1 - y_2$ changes linearly in time — no quadratic, no square roots. Gravity is invisible to the gap.

The master solution pattern

Route one — positions equal. Write $y_1(t)$ and $y_2(t)$ from equation (1), set $y_1 = y_2$ , solve for $t$ , substitute back for the height. Mechanical, safe, and what CBSE/ISC structured questions expect shown. Route two — the relative shortcut. The gap closes at the constant rate $|u_1 - u_2|$ (relative speed at launch), so

t_{\text{meet}} = \frac{\text{initial gap}}{\text{relative speed}} \qquad (2)

where: the initial gap is the launch separation and the relative speed is $|u_1 - u_2|$ — with simultaneous launches only; delayed launches need the gap and relative speed evaluated at the second launch. This is a one-line answer to what looks like a quadratic problem, and it is the difference between 90 seconds and 20 seconds on a JEE-Main MCQ.

situation	governing result	meeting/answer
drop from $H$ + throw up $u$ (simultaneous)	gap $H$ closes at rate $u$	$t = H/u$ ; height $= H - \tfrac{1}{2}g(H/u)^2$
meet in the air at all?	dropped ball lands at $\sqrt{2H/g}$	need $u \ge \sqrt{gH/2}$
two equal throws $u$ , delay $\Delta t$	symmetry about apexes	meet at $t = \dfrac{u}{g} + \dfrac{\Delta t}{2}$ from first throw
packet from balloon rising at $u$ , height $h$	$-h = ut - \tfrac{1}{2}gt^2$	$t = \dfrac{u + \sqrt{u^2 + 2gh}}{g}$
speed symmetry	up-speed at a height $=$ down-speed there	free sanity check

On a position–time graph the story is one picture: two parabolas of the same curvature, whose vertical gap shrinks along a straight line.

Figure

Position-time graph of the tower problem — two parabolas, one crossing

Derivation ATower meeting by both routes

We solve the defining problem of the archetype — ball 1 dropped from height $H$ at the same instant ball 2 is thrown up from the ground at speed $u$ — for the meeting time and height, first by setting position functions equal (the board-marks route), then by the relative-velocity shortcut (the exam-speed route). Watching both routes land on the same answer is the fastest way to trust the shortcut. Keep up as positive throughout; the dropped ball's initial velocity is zero, not $-g$ .

Step 1. Position functions on one axis and clock:

y_1(t) = H - \tfrac{1}{2}g t^2, \qquad y_2(t) = u\,t - \tfrac{1}{2}g t^2

Why: both bodies carry the identical $-\tfrac{1}{2}gt^2$ term — same $g$ , same clock — which is exactly what will cancel.

Step 2. Set $y_1 = y_2$ :

H - \tfrac{1}{2}gt^2 = ut - \tfrac{1}{2}gt^2 \;\;\Rightarrow\;\; H = ut \;\;\Rightarrow\;\; t_{\text{meet}} = \frac{H}{u}

Why: the quadratic terms cancelled BEFORE any algebra — the meeting condition was secretly linear all along. Gravity delays both balls equally, so it cannot influence when they meet.

Step 3. Meeting height by substituting into either function:

y_{\text{meet}} = u\left(\frac{H}{u}\right) - \frac{1}{2}g\left(\frac{H}{u}\right)^2 = H - \frac{gH^2}{2u^2}

Step 4. The shortcut route: relative to ball 2, ball 1 has zero relative acceleration and closing speed $|0 - u| = u$ ; the gap is $H$ . So $t_{\text{meet}} = H/u$ — one line, same answer.

Step 5. Reality check — do they meet in the air? The dropped ball lands at $t = \sqrt{2H/g}$ ; require $H/u \le \sqrt{2H/g}$ , i.e. $u \ge \sqrt{gH/2}$ . Slower throws meet the dropped ball only "below the ground" — meaning never. Verified: $H = 100\ \mathrm{m}$ , $u = 25\ \mathrm{m/s}$ , $g = 9.8$ : $t = 4\ \mathrm{s}$ , height $= 100 - 78.4 = 21.6\ \mathrm{m}$ ; threshold $\sqrt{9.8 \times 50} = 22.1\ \mathrm{m/s} < 25$ ✓ they meet in the air. Both position functions give $21.6\ \mathrm{m}$ .

Go deeper · Derivation B · 1 minThe balloon packet and its inherited velocity

Derivation BThe balloon packet and its inherited velocity

We derive the flight time of a packet released from a balloon that is rising at speed $u$ when the packet is let go at height $h$ . The physical crux sits in the initial condition: "released" means the packet stops being held, not that it stops moving — it leaves with the balloon's own $+u$ . So the packet first climbs, decelerating at $g$ , then falls past the release height and down to the ground. The algebra is one quadratic; choosing the correct root is part of the derivation.

Step 1. Up positive, origin at the release point, clock starting at release. The packet's displacement:

s(t) = u\,t - \tfrac{1}{2}g t^2

Why: $u$ is positive — the packet inherits the balloon's upward velocity. Writing $u = 0$ here is THE classic error; the packet is not "dropped from rest" in any frame fixed to the ground.

Step 2. The ground lies a distance $h$ below the origin, so impact is $s = -h$ :

-h = ut - \tfrac{1}{2}gt^2 \;\;\Rightarrow\;\; \tfrac{1}{2}gt^2 - ut - h = 0

Step 3. Solve the quadratic for $t$ :

t = \frac{u + \sqrt{u^2 + 2gh}}{g}

Why: the discriminant $u^2 + 2gh$ is always positive (the packet always lands), and we keep the $+$ root because the $-$ root is negative — a fictitious impact before release.

Step 4. Impact velocity from $v = u - gt$ : substituting the flight time gives $v = -\sqrt{u^2 + 2gh}$ — downward, with a magnitude you could also read directly from $v^2 = u^2 + 2gh$ (displacement $-h$ , the $u^2$ appearing because speed at the release height is $u$ both going up and coming down). Verified: $u = 5\ \mathrm{m/s}$ , $h = 60\ \mathrm{m}$ , $g = 10$ : $t = (5 + \sqrt{25 + 1200})/10 = (5 + 35)/10 = 4.0\ \mathrm{s}$ ; impact speed $\sqrt{1225} = 35\ \mathrm{m/s}$ ; check by $v = u - gt = 5 - 40 = -35\ \mathrm{m/s}$ ✓.

Go deeper · Derivation C · 1 minWhy the gap between two free bodies closes at constant speed

Derivation CWhy the gap between two free bodies closes at constant speed

We prove the archetype's master theorem: for ANY two bodies in free vertical flight — whatever their launch heights, speeds and directions — the separation changes linearly with time. This is the licence behind every "relative velocity" one-liner in this family, so it deserves its own three lines. The strategy is bare subtraction of position functions; the physics is that the $g$ -terms are identical.

Step 1. Write both position functions on the shared axis and clock:

y_1(t) = y_{01} + u_1 t - \tfrac{1}{2}gt^2, \qquad y_2(t) = y_{02} + u_2 t - \tfrac{1}{2}gt^2

Step 2. Subtract:

\Delta y(t) = y_1 - y_2 = (y_{01} - y_{02}) + (u_1 - u_2)\,t

Why: the $-\tfrac{1}{2}gt^2$ terms cancel because $g$ is the same for both bodies — Galileo's equal-fall fact doing algebraic work. Nothing quadratic survives.

Step 3. Read the result: the gap is a straight-line function of time with slope $u_1 - u_2$ . Bodies meet when $\Delta y = 0$ , i.e. after time $|\,y_{01} - y_{02}|/|u_1 - u_2|$ (simultaneous launches). If $u_1 = u_2$ (the juggler's equal throws), the gap never changes while both are in flight — they cross only if a delay built the gap differently. Verified: for the tower problem, $\Delta y(0) = H$ , slope $= 0 - u = -u$ , zero at $t = H/u$ — reproducing Derivation A. For two equal throws $\Delta t$ apart (both at $u = 20$ , $g = 10$ ), positions coincide at $t = u/g + \Delta t/2 = 2.5\ \mathrm{s}$ , matching the worked example below.

The variants examiners actually pose

Drop + throw (the tower meeting). Solved above; the exam dresses it as police towers, wells, cliffs. Watch for the condition add-on: "if they meet at the highest point of ball 2's flight" forces $t_{\text{meet}} = u/g$ , and then $H = ut = u^2/g$ — with the elegant consequence that they meet at $u^2/2g = H/2$ , exactly half the tower height, always.

The juggler (equal throws, delayed). Both balls thrown up at $u$ , the second $\Delta t$ later. By time symmetry the crossing sits halfway between the apexes: $t = u/g + \Delta t/2$ , the balls passing at equal and opposite speeds $g\,\Delta t/2$ .

The balloon packet. Inherited $+u$ ; solved above. NEET's favourite twist: "how much higher does the packet rise after release?" — answer $u^2/2g$ , from $v = 0$ .

Ball past a window / nth-second distances. One body, same machinery: displacement in the $n$ -th second is $u - g(n - \tfrac{1}{2})$ , and window-transit times pin down launch speeds.

Limiting cases

case	result	meaning
$u \to \infty$ (tower problem)	$t = H/u \to 0$ , meet at height $\to H$	a bullet-fast throw meets the ball where it was dropped
$u = \sqrt{gH/2}$	meet exactly at the ground	slowest throw that still "meets"; slower never does
balloon $u = 0$	$t = \sqrt{2h/g}$	packet reduces to a simple drop — formula self-checks
juggler $\Delta t \to 0$	balls coincide throughout	zero delay = same launch = same flight
equal throws, both in flight	gap constant	$u_1 = u_2$ kills the closing speed entirely

Common trap“A packet released from a rising balloon immediately starts falling”

It cannot — at the instant of release the packet has the balloon's velocity, $+u$ upward, and gravity needs time $u/g$ to eat that velocity before the packet even begins to descend. It climbs a further $u^2/2g$ above the release point first. For a balloon rising at $5\ \mathrm{m/s}$ ( $g = 10$ ): the packet rises $1.25\ \mathrm{m}$ more for $0.5\ \mathrm{s}$ , then falls. Every option list for this setup contains the "dropped from rest" answer — for the $60\ \mathrm{m}$ balloon that wrong route gives $t = \sqrt{12} = 3.46\ \mathrm{s}$ instead of the true $4.0\ \mathrm{s}$ . The word "released" or "dropped" never means $u = 0$ when the platform is moving.

Worked examples

1 · Board level — the tower meeting (CBSE/ISC pattern, 3 marks)

A stone is dropped from the top of a $45\ \mathrm{m}$ tower at the same instant a ball is thrown vertically upward from the ground at $25\ \mathrm{m/s}$ , along the same vertical line. Take $g = 10\ \mathrm{m/s^2}$ . Find when and where they meet.

Step 1. Declare the convention (this earns a method mark): up positive, origin at the ground, $t = 0$ at release. Stone: $y_1 = 45 - 5t^2$ . Ball: $y_2 = 25t - 5t^2$ .

Step 2. Meeting condition $y_1 = y_2$ : the $-5t^2$ terms cancel, leaving $45 = 25t \Rightarrow t = \boxed{1.8\ \mathrm{s}}$ .

Step 3. Height: $y_2 = 25(1.8) - 5(1.8)^2 = 45 - 16.2 = \boxed{28.8\ \mathrm{m}}$ above the ground.

Step 4. Sanity: the stone has fallen $5(1.8)^2 = 16.2\ \mathrm{m}$ from $45\ \mathrm{m}$ — also $28.8\ \mathrm{m}$ ✓. And $u = 25 > \sqrt{gH/2} = \sqrt{225} = 15\ \mathrm{m/s}$ , so an in-air meeting was guaranteed.

2 · Board level — the juggler's two balls (ISC/CBSE pattern)

A juggler throws a ball vertically upward at $20\ \mathrm{m/s}$ . Exactly $1.0\ \mathrm{s}$ later he throws a second ball upward at the same speed from the same point. Take $g = 10\ \mathrm{m/s^2}$ . When and where do the balls cross?

Step 1. One clock, started at the first throw. Ball A: $y_A = 20t - 5t^2$ . Ball B (aloft for $t - 1$ ): $y_B = 20(t-1) - 5(t-1)^2$ , valid for $t \ge 1$ .

Step 2. Set equal: $20t - 5t^2 = 20(t-1) - 5(t-1)^2$ . Expand the right side: $20t - 20 - 5t^2 + 10t - 5$ . The $20t$ and $-5t^2$ cancel from both sides, leaving $0 = -25 + 10t \Rightarrow t = \boxed{2.5\ \mathrm{s}}$ .

Step 3. Height: $y_A = 20(2.5) - 5(6.25) = 50 - 31.25 = \boxed{18.75\ \mathrm{m}}$ .

Step 4. Read the symmetry: A peaks at $t = 2\ \mathrm{s}$ , B peaks at $t = 3\ \mathrm{s}$ — the crossing sits exactly halfway between the apexes, at $t = u/g + \Delta t/2$ . A is falling at $g(t - 2) = 5\ \mathrm{m/s}$ while B rises at $5\ \mathrm{m/s}$ : equal and opposite, as the time symmetry demands.

3 · Competitive level — NEET/JEE-Main pattern: the balloon packet (~3 minutes)

A balloon rises steadily at $5\ \mathrm{m/s}$ . When it is $60\ \mathrm{m}$ above the ground, a packet slips from it. Take $g = 10\ \mathrm{m/s^2}$ . (a) How long does the packet take to reach the ground? (b) With what speed does it land? (c) What maximum height above the ground does it reach?

Step 1. Initial velocity is the balloon's: $u = +5\ \mathrm{m/s}$ (up positive, origin at release). Displacement to the ground: $s = -60\ \mathrm{m}$ .

Step 2. $-60 = 5t - 5t^2 \Rightarrow t^2 - t - 12 = 0 \Rightarrow (t-4)(t+3) = 0 \Rightarrow t = \boxed{4.0\ \mathrm{s}}$ (discard $-3$ ).

Step 3. Landing velocity: $v = u - gt = 5 - 40 = -35 \Rightarrow$ speed $\boxed{35\ \mathrm{m/s}}$ downward. (Check: $v^2 = 25 + 2(10)(60) = 1225$ ✓.)

Step 4. Extra climb: $u^2/2g = 25/20 = 1.25\ \mathrm{m}$ , so maximum height $= 60 + 1.25 = \boxed{61.25\ \mathrm{m}}$ , reached at $t = 0.5\ \mathrm{s}$ . Timing note: ~3 min, and almost all of it is part (a)'s quadratic — which factorizes if you divide through by $5$ first. The mark-scheme trap is $u = 0$ : that gives $t = \sqrt{12} = 3.46\ \mathrm{s}$ and $34.6\ \mathrm{m/s}$ , both sitting in the options waiting. The word "slips" does not mean "starts from rest".

4 · Competitive level — JEE-Advanced stretch: meet at the apex (~4 minutes)

A ball is dropped from the top of a tower of height $H$ at the instant a second ball is thrown up from its base at $u = 20\ \mathrm{m/s}$ . They collide exactly when the second ball is at the top of its flight. Take $g = 10\ \mathrm{m/s^2}$ . Find $H$ , the time of collision, and the height at which they collide — and show the collision height is $H/2$ regardless of $u$ .

Step 1. "At the top of its flight" pins the time: the thrown ball peaks at $t = u/g = 20/10 = 2.0\ \mathrm{s}$ . So $t_{\text{meet}} = \boxed{2.0\ \mathrm{s}}$ .

Step 2. The tower meeting always satisfies $H = u\,t_{\text{meet}}$ (Derivation A). Hence $H = 20 \times 2.0 = \boxed{40\ \mathrm{m}}$ — equivalently $H = u^2/g$ .

Step 3. Collision height = the thrown ball's apex: $u^2/2g = 400/20 = \boxed{20\ \mathrm{m}}$ .

Step 4. Generality: $\dfrac{\text{collision height}}{H} = \dfrac{u^2/2g}{u^2/g} = \dfrac{1}{2}$ — the apex-meeting always happens at half the tower height, whatever $u$ is. Cross-check: the dropped ball falls $\tfrac{1}{2}g t^2 = 5(4) = 20\ \mathrm{m}$ from $40\ \mathrm{m}$ ✓. Timing note: ~4 min if you derive; ~90 s if you carry the pattern " apex meeting $\Rightarrow H = u^2/g$ , collide at $H/2$ " as a known result. JEE-Advanced loves asking for the ratio precisely because the $u$ -dependence cancels — an option like "depends on u" is the examiner testing whether you did the algebra.

Exam traps

Inherited velocity ignored. Anything released from a moving platform — balloon, lift, rising drone — starts with the platform's velocity. "Dropped" describes the release, not the initial velocity. This single trap decides more balloon-packet marks than all the algebra combined.

Two private clocks. In delayed-launch problems, write the second body's position with $(t - t_0)$ , never a fresh $t$ . Expanding $-\tfrac{1}{2}g(t - t_0)^2$ carelessly — losing the cross term — is the standard juggler error.

Sign convention drift. One body solved with up positive, the other with down positive, then positions "set equal": guaranteed nonsense. Declare the convention once, at the top, and force every $u$ and every $g$ through it.

Meeting "below the ground". The algebra happily returns meeting times after one body has already landed. Always check $t_{\text{meet}}$ against each body's landing time — in the tower problem, that is the $u \ge \sqrt{gH/2}$ condition.

Distance vs displacement. A ball up-and-back-down has displacement zero but distance $2 \times$ (max height). "Distance travelled in the 3rd second" around the apex must be split at the turning point — the naive formula gives displacement, not distance.

"Acceleration is zero at the top." Velocity is zero at the apex; acceleration is $-g$ throughout, apex included. Any meeting-at-apex analysis that switches off gravity for an instant collapses.

Which $g$ ? CBSE/ISC expect $g = 9.8\ \mathrm{m/s^2}$ unless stated; JEE-Main/Advanced and NEET almost always declare $g = 10$ in the question. Use what is stated, and never mix values between the two bodies — they share one $g$ by definition.

Calculator habit. Divide quadratics like $t^2 - t - 12 = 0$ by common factors first — factorable integers are a deliberate gift, and the quadratic formula under time pressure breeds sign slips.

Origins · 1 minGalileo Galilei and how this was discovered

Galileo Galilei

1564–1642 · Pisa, Padua, Florence

Galileo turned falling into an exact science. Aristotle had taught that heavy bodies fall faster in proportion to their weight; Galileo — by argument (tie a light ball to a heavy one: must the pair fall faster or slower than the heavy one alone?) and by experiment with inclined planes that "diluted" gravity — established that all bodies share one acceleration. His 1638 Two New Sciences states the odd-number law of distances and the time-squared rule, the DNA of every kinematic equation in this article. The Pisa drop may be legend, but in 1971 Apollo 15 commander David Scott performed it on the Moon with a hammer and a falcon feather: they struck the lunar dust together, on camera. Every "two bodies under gravity" problem is Galileo's equal-fall principle wearing an exam costume.

History

1589

— Legend places Galileo at the Leaning Tower of Pisa, dropping unequal weights that land together.

1604

— Galileo works out the times-squared law and the odd-number rule for distances in successive intervals.

1638

— Two New Sciences publishes the mathematics of uniformly accelerated fall — the foundation of this whole archetype.

1687

— Newton's Principia explains why all bodies share one

g

: gravity pulls in proportion to the very mass that resists it.

1971

— Apollo 15's hammer-and-feather drop shows equal fall in the Moon's vacuum to a live television audience.

Going further

Texts. NCERT Physics Class 11, Ch. 3 (motion under gravity) · HC Verma, Concepts of Physics Vol. 1, Ch. 3 · Resnick–Halliday–Walker, Ch. 2.

Provenance. All results are standard kinematics. A verifier should recompute: tower meeting $H = 100$ , $u = 25$ , $g = 9.8$ : $t = 4\ \mathrm{s}$ , height $21.6\ \mathrm{m}$ , threshold $22.1\ \mathrm{m/s}$ · Example 1 ( $H = 45$ , $u = 25$ , $g = 10$ ): $t = 1.8\ \mathrm{s}$ , $28.8\ \mathrm{m}$ · Example 2 ( $u = 20$ , $\Delta t = 1$ , $g = 10$ ): $t = 2.5\ \mathrm{s}$ , $18.75\ \mathrm{m}$ , crossing speeds $\pm 5\ \mathrm{m/s}$ · Example 3 ( $u = 5$ , $h = 60$ , $g = 10$ ): $t = 4.0\ \mathrm{s}$ , $35\ \mathrm{m/s}$ , max $61.25\ \mathrm{m}$ · Example 4 ( $u = 20$ , $g = 10$ ): $H = 40\ \mathrm{m}$ , collide $t = 2\ \mathrm{s}$ at $20\ \mathrm{m} = H/2$ · balloon warning-box wrong-route $t = \sqrt{12} = 3.46\ \mathrm{s}$ .

Prerequisites. Free Fall & Motion Under Gravity · Kinematic Equations for Uniform Acceleration. Leads to. Relative Velocity & Reference Frames (the shortcut, generalized to 2D) · Chase, Catch & Meet (the same machine laid horizontal).

Why is the relative acceleration of two falling bodies exactly zero?

Because gravity gives every free body the same acceleration $-g$ , independent of its mass, velocity or direction of motion — Galileo's equal-fall fact. Subtracting the two acceleration vectors: $(-g) - (-g) = 0$ . The gap between the bodies then behaves like a problem with no gravity at all: it just changes at the constant rate set by the launch velocities. That's why meeting times come out of linear equations even though each body separately moves on a parabola in time.

Doesn't the heavier body fall at least a little faster?

In vacuum, no — mass cancels exactly, because the gravitational force on a body is proportional to the same mass that resists acceleration ( $F = mg = ma \Rightarrow a = g$ ). In air, drag muddies things: a feather falls slower than a hammer on Earth, but that is air resistance, not gravity playing favourites. Exam problems say "neglect air resistance" precisely to hand you the clean equal- $g$ world; the Apollo 15 hammer-and-feather drop shows that world is real.

How do I know whether the two bodies actually meet in the air?

Compute the meeting time, then check both bodies are still airborne then. In the drop-plus-throw problem: $t_{\text{meet}} = H/u$ must not exceed the dropped ball's landing time $\sqrt{2H/g}$ , which rearranges to $u \ge \sqrt{gH/2}$ . If the throw is slower than that, the "meeting" the algebra reports happens at negative height — after the stone has already hit the ground. Always translate the algebraic answer back into the physical timeline.

In delayed-launch problems, why (t − t₀) and not a new clock for the second body?

You can run a private clock for each body — but then the meeting condition mixes two different time variables and you must add the offset anyway, usually incorrectly under pressure. One master clock with $(t - t_0)$ for the latecomer keeps every position function comparable: setting $y_1(t) = y_2(t)$ is legitimate because both are evaluated at the same instant. Just remember the second body's function is only valid for $t \ge t_0$ ; before that it hasn't been launched.

Why does the packet from the balloon rise first instead of falling immediately?

Velocity doesn't change instantly — only forces change it, at a finite rate. At release the packet moves upward at the balloon's $5\ \mathrm{m/s}$ ; gravity then removes $10\ \mathrm{m/s}$ of upward velocity every second, so for half a second the packet is still climbing (another $u^2/2g = 1.25\ \mathrm{m}$ ). Watch videos of skydivers leaving a rising hot-air balloon: they visibly float upward relative to the ground for a moment. "Released" describes the grip, not the motion.

Can I take downward as positive instead?

Absolutely — sign conventions are choices, not laws. Downward-positive is cleaner when everything falls; up-positive is cleaner the moment anything moves upward (like the balloon packet's inherited velocity). The only crime is switching conventions mid-problem or between the two bodies. Declare once, apply it to every $u$ , $g$ and $s$ , and either convention gives identical physical answers.

What's the fastest way to sanity-check a meeting answer in the exam?

Three free checks. First, substitute $t_{\text{meet}}$ into BOTH position functions — they must agree (Example 1: both give $28.8\ \mathrm{m}$ ). Second, timeline: is $t_{\text{meet}}$ before both landing times? Third, symmetry: speeds at a shared height match up/down (Example 2's balls cross at equal and opposite $5\ \mathrm{m/s}$ ). Ten seconds of checking catches the sign and clock errors that produce plausible-looking wrong answers.