ARCHETYPE: Projectile from a Height

Roll a marble off the edge of a table and, at the same instant, drop a second marble from the same height. Which hits the floor first? Almost everyone says the dropped one — the fired marble has "so much further to travel". The floor disagrees: click, one sound, two landings, same moment. The marble racing sideways at full speed falls in exactly the time the dropped one does, because horizontal motion has no vote in vertical affairs. That one demonstration is the engine of this archetype. A projectile launched from a height — a ball off a cliff, a package from an aircraft, a stone from a tower — splits into two strangers who share only a clock: uniform motion sideways, free fall downwards. The fall time comes from the height alone, $t = \sqrt{2h/g}$ ; the launch speed decides where it lands, not when. Every board tests this setup — CBSE and ICSE as a suvat exercise, JEE and NEET with landing velocities and angled launches, AP, IB and A-Level as the definitive test of whether you truly believe in independence of components. So treat the setup as the concept: one master pattern (measure everything downward from the launch point), formulas that fall out in two lines, the angled-from-a-height upgrade with its quadratic and one honest root, and the traps — this archetype has a famous one, and it involves trusting intuition instead of the clock.

Key takeaways

A projectile launched horizontally from height $h$ falls for $t = \sqrt{2h/g}$ — independent of launch speed $u$ .
Range from the base: $R = ut = u\sqrt{2h/g}$ ; doubling $u$ doubles $R$ but leaves the flight time untouched.
Landing velocity: $v_x = u$ stays constant, $v_y = gt = \sqrt{2gh}$ grows from zero; magnitude $v = \sqrt{u^2 + 2gh}$ , direction $\tan\phi = v_y/v_x$ below horizontal.
Measure $y$ downward from the launch point and every sign problem dies: $y = \tfrac12 gt^2$ for horizontal launch.
Angled launches from a height give a quadratic in $t$ — keep the positive root, the negative one landed before you threw.
Key idea: height alone sets the clock; horizontal speed only decides where the projectile is when the clock runs out.

The core insight

For any projectile from a height, the vertical motion is pure free fall that fixes the flight time from $h$ and $g$ alone, while the horizontal velocity — constant throughout — simply multiplies that time into a landing distance.

Where this lives today

Every ski jumper leaving the takeoff table, every crate leaving a warehouse chute is a projectile from a height — engineers still size landing zones with $u\sqrt{2h/g}$ before any simulation runs. Humanitarian air-drop planners solve Example 3 for real: release point = target minus $u\sqrt{2h/g}$ , corrected for wind and parachute drag. Motocross course designers shape landing ramps to meet the rider's parabola tangentially — matching the landing angle $\tan\phi = \sqrt{2gh}/u$ so the impact component vanishes. And 400 km up, the International Space Station runs the archetype's limiting case continuously: horizontal speed so large that the fall — genuine, permanent free fall — never finds the ground. That story, with its energies and periods, is the Satellites module's opening move.

The setup

Geometry first, because in an archetype the geometry is the physics. A projectile leaves the edge of a table, cliff or tower of height $h$ with a purely horizontal velocity $u$ . From that instant only gravity acts. Two things happen simultaneously and independently: the projectile drifts sideways at the unchanging $u$ , and it falls exactly like a dropped stone — $5, 20, 45$ metres after $1, 2, 3$ seconds (at $g = 10\ \mathrm{m/s^2}$ ), however fast it moves sideways.

Figure

Projectile from a height — the dropped and the fired object fall together

One convention kills nearly every sign error in this family: measure the vertical coordinate downward from the launch point. Down is where everything happens — displacement, gravity, the growing velocity. With down positive, a horizontal launch has zero initial vertical velocity, acceleration $+g$ , and the vertical suvat equations from Kinematic Equations carry no minus signs at all. (For launches angled upward, keep down positive and give the initial vertical velocity one minus sign, stated once.) The licence for splitting the motion is that gravity has no horizontal component: Newton's second law runs separately per direction — "no force, no change" horizontally, free fall vertically. The components share nothing except time, which is exactly why finding the flight time is always move one.

The master solution pattern

Horizontal direction — uniform motion at the launch speed:

x = u\,t \qquad (1)

where: $x$ — horizontal distance from the base of the launch point; $u$ — launch speed. There is nothing else: no acceleration, no $t^2$ , no drama. If a question asks only about horizontal distance, the entire difficulty must be hiding in finding $t$ . Vertical direction — free fall from rest (down positive):

y = \tfrac{1}{2} g t^2 \qquad (2)

where: $y$ — distance fallen since launch. This is a dropped stone's equation, untouched by $u$ . Setting $y = h$ gives the archetype's heartbeat, the flight time:

t = \sqrt{\frac{2h}{g}} \qquad (3)

where: $h$ — the launch height. Read it physically: the clock is set by the height alone. A bullet fired horizontally at 300 m/s from shoulder height and a coin nudged off the same shoulder land at the same instant. Doubling the height multiplies the time by $\sqrt{2}$ , not 2. Substituting (3) into (1) gives the range from the base:

R = u\sqrt{\frac{2h}{g}} \qquad (4)

where: $R$ — the landing distance from the foot of the cliff. Range is linear in $u$ (double the speed, land twice as far) but only square-root in $h$ . Finally, the landing velocity assembles itself from the two components:

v_x = u, \qquad v_y = gt = \sqrt{2gh}, \qquad v = \sqrt{u^2 + 2gh} \qquad (5)

where: $v$ — landing speed, at angle $\tan\phi = v_y/v_x$ below the horizontal. The $v^2 = u^2 + 2gh$ shape is no accident — it is the work–energy bookkeeping in kinematic disguise, a bridge you will cross properly in the Kinetic Energy & the Work–Energy Theorem module.

quantity	horizontal launch from height $h$	where it comes from
flight time	$t = \sqrt{2h/g}$	vertical free fall alone
range from base	$R = u\sqrt{2h/g}$	$x = ut$ at the fall time
landing $v_y$	$\sqrt{2gh}$	$v_y = gt$ or $v_y^2 = 2gh$
landing speed	$\sqrt{u^2 + 2gh}$	components in quadrature
landing angle below horizontal	$\tan\phi = \sqrt{2gh}/u$	ratio of components
trajectory shape	$y = gx^2/(2u^2)$	eliminate $t$ between (1) and (2)

Derivation AFlight time and range — two lines once the axes are honest

We derive the flight time (3) and range (4) for a horizontal launch from height $h$ — the results every marking scheme wants stated, then used. The strategy is the standard two-step of this archetype: solve the vertical problem alone for the time, then feed that time to the horizontal equation. The only decision worth a sentence is the axis convention: down positive, origin at the launch point, so the initial vertical velocity is zero and no sign ever flips.

Step 1. Set up. Origin at the launch point, $x$ horizontal along the throw, $y$ positive downward. Initial velocities: $u_x = u$ , $u_y = 0$ . Accelerations: $a_x = 0$ , $a_y = +g$ . Why: with down positive, the fall is a positive displacement under positive acceleration from zero velocity — the simplest suvat sentence that exists.

Step 2. Vertical clock. From $y = u_y t + \tfrac12 g t^2$ with $u_y = 0$ : the projectile reaches the ground when $\tfrac12 g t^2 = h$ , so

t = \sqrt{\frac{2h}{g}}

Step 3. Horizontal ruler. Uniform motion the whole flight:

R = u\,t = u\sqrt{\frac{2h}{g}}

Why: the two steps cannot be merged or reordered — time is manufactured by the vertical motion and spent by the horizontal one. Every "projectile from a height" question, however decorated, is these two steps plus reading comprehension.

Step 4. Landing velocity. $v_x = u$ (never changed); $v_y = gt = \sqrt{2gh}$ ; combine by Pythagoras for the speed and $\tan\phi = v_y/v_x$ for the direction below horizontal. Verified: $h = 45$ m, $u = 20$ m/s, $g = 10$ : $t = \sqrt{9} = 3$ s, $R = 60$ m, $v_y = 30$ m/s, $v = \sqrt{400+900} = \sqrt{1300} \approx 36$ m/s at $\tan\phi = 1.5$ . Limits: $u \to 0$ reduces to a pure drop ( $R \to 0$ , $t$ unchanged); $h \to 0$ gives $t \to 0$ — no cliff, no flight.

Go deeper · Derivation B · 1 minThe trajectory is a parabola — and the two-marbles demo hides inside it

Derivation BThe trajectory is a parabola — and the two-marbles demo hides inside it

We derive the shape of the path, $y(x)$ , by eliminating time between the horizontal and vertical equations, and read two exam-grade facts out of the result. The strategy is substitution: equation (1) makes $t$ a proxy for $x$ , so free fall in time becomes a parabola in space. Watch how the launch speed ends up flattening the parabola without changing the fall.

Step 1. From $x = ut$ , express the shared clock as $t = x/u$ .

Step 2. Substitute into $y = \tfrac12 gt^2$ :

y = \frac{g}{2u^2}\,x^2

where $y$ is measured downward from the launch point. The path is a half-parabola opening downward, starting horizontal at the edge. Why: the curve starts flat because $v_y = 0$ at launch — a projectile from a height always leaves along the horizontal it was given, then bends. A faster launch (larger $u$ ) shrinks the coefficient $g/2u^2$ : same fall, stretched over more horizontal distance — a flatter arc, not a slower one.

Step 3. Read off the demo. Two objects leave the same edge at the same moment, one with $u = 0$ (dropped) and one with $u > 0$ (fired). Both obey $y = \tfrac12 gt^2$ — identical vertical histories, so equal heights at every instant and one shared landing time $\sqrt{2h/g}$ . The fired object simply visits its heights somewhere else.

Step 4. Cross-check against the angled-launch trajectory from Projectile Motion, $y_{\text{up}} = x\tan\theta - \tfrac{g x^2}{2u^2\cos^2\theta}$ : set $\theta = 0$ and flip the sign convention (their $y$ is up) to recover Step 2 exactly — the horizontal launch is the $\theta = 0$ member of the family. Verified: at $u = 20$ m/s, $g = 10$ : $y(20\ \mathrm{m}) = \tfrac{10}{2(400)}(400) = 5$ m — the same 5 m any dropped object falls in its first second, and indeed $x = 20$ m is reached at $t = 1$ s. Parabola confirmed point by point.

Go deeper · Derivation C · 1 minAngled launch from a height — one quadratic, one honest root

Derivation CAngled launch from a height — one quadratic, one honest root

We derive the flight time when the projectile leaves the cliff-top at speed $u_0$ at an angle $\theta$ above the horizontal — the JEE/A-Level upgrade of the archetype. The strategy: same axes (down positive), but now the initial vertical velocity is upward, so it enters with a minus sign; the landing condition becomes a quadratic in $t$ whose two roots both mean something, but only one of them is your landing.

Step 1. Components at launch: $u_x = u_0\cos\theta$ (constant forever), $u_y = -u_0\sin\theta$ (minus because up, with down positive).

Step 2. Landing condition — total downward displacement equals the height:

h = -u_0\sin\theta\; t + \tfrac12 g t^2

Step 3. Solve the quadratic $\tfrac12 g t^2 - u_0\sin\theta\, t - h = 0$ :

t = \frac{u_0\sin\theta + \sqrt{u_0^2\sin^2\theta + 2gh}}{g}

Why: the discriminant $u_0^2\sin^2\theta + 2gh$ is always positive — a launched projectile always lands. The quadratic's other root is negative: it is the fictitious earlier moment at which a projectile on this same parabola would have left the ground. Real clocks start at launch, so keep the positive root.

Step 4. Then range $= u_0\cos\theta \times t$ , and the landing $v_y = -u_0\sin\theta + gt$ , exactly as before. For a launch angled below the horizontal, flip one sign: $u_y = +u_0\sin\theta$ , and the quadratic gives a shorter time — the projectile is helped downward. Verified: $u_0 = 25$ m/s, $\theta = 37°$ ( $\sin 37° = 0.6$ , $\cos 37° = 0.8$ ), $h = 20$ m, $g = 10$ : $t = \tfrac{15 + \sqrt{225 + 400}}{10} = \tfrac{15+25}{10} = 4$ s; check in Step 2: $-15(4) + 5(16) = 20$ ✓. Range $= 20 \times 4 = 80$ m. Setting $\theta = 0$ collapses the formula to $\sqrt{2h/g}$ ✓.

Figure

Angled launch from a cliff — one parabola, one honest root, and the landing triangle

The variants examiners actually pose

The package from the plane. An aircraft flying horizontally at speed $u$ and height $h$ releases a package. The package starts with the plane's velocity — it is a horizontal launch with that $u$ . Favourite twist: where is the plane when the package lands? Directly above the package (both kept the same $v_x$ ; ignore air resistance). The package traces the parabola; the pilot watches it fall straight down beneath them.

Landing conditions. "Does the stone clear the 70 m wide moat?" — compute $R$ and compare. "The ball lands on a terrace 20 m below the roof" — set $y = 20$ m, not the full height. The landing surface defines $y$ ; read the geometry before computing.

Hit a given point. To land at horizontal distance $d$ on ground $h$ below: from (4), $u = d\sqrt{g/2h}$ — the required launch speed. One unknown, one equation; this inversion is a one-mark gift dressed as design engineering.

Velocity direction mid-flight. "When does the velocity make $45°$ with the horizontal?" — when $v_y = v_x$ , i.e. $gt = u$ , at $t = u/g$ . No positions needed: direction questions are ratio questions.

Limiting cases

case	result	meaning
$u \to 0$	$t = \sqrt{2h/g}$ , $R \to 0$	pure drop: the archetype contains free fall as its zero-speed member
$u$ very large	$R \propto u$ , same $t$	flatter, longer arc — but the floor arrives on the same schedule
$h \to 0$	$t \to 0$ , $R \to 0$	no height, no flight — a ground-level horizontal "launch" lands instantly
$\theta \to 0$ in the angled formula	$t \to \sqrt{2h/g}$	the quadratic collapses to the horizontal-launch clock
$g$ larger (another planet)	$t$ and $R$ shrink as $1/\sqrt{g}$	same throw on the Moon ( $g/6$ ): $\sqrt{6} \approx 2.4$ times the time and range

Worked examples

1 · Board level — off the cliff (CBSE/ICSE pattern, 3 marks)

A stone is thrown horizontally at $20\ \mathrm{m/s}$ from a cliff $45\ \mathrm{m}$ high. Take $g = 10\ \mathrm{m/s^2}$ . Find (a) the time of flight, (b) the distance from the foot of the cliff at which it lands, (c) the speed with which it strikes the ground.

Step 1. Time from the vertical fall alone: $t = \sqrt{2h/g} = \sqrt{2(45)/10} = \sqrt{9} = \boxed{3.0\ \mathrm{s}}$ .

Step 2. Range from the base: $R = ut = 20 \times 3 = \boxed{60\ \mathrm{m}}$ .

Step 3. Landing components: $v_x = 20\ \mathrm{m/s}$ , $v_y = gt = 30\ \mathrm{m/s}$ . Speed $v = \sqrt{20^2 + 30^2} = \sqrt{1300} = \boxed{36\ \mathrm{m/s}}$ (2 s.f.), at $\tan\phi = 30/20 = 1.5$ , i.e. $\phi \approx 56°$ below horizontal. Sanity: $v$ must exceed both components individually and beat the launch speed ✓ (falling adds energy, never removes it).

2 · Board level — the table-edge marble (A-Level/AP pattern)

A marble rolls off a table of height $1.25\ \mathrm{m}$ at $2.0\ \mathrm{m/s}$ . With $g = 10\ \mathrm{m/s^2}$ (stated by the paper), find where it lands and sketch what a second, simultaneously dropped marble does.

Step 1. Fall time: $t = \sqrt{2(1.25)/10} = \sqrt{0.25} = 0.50\ \mathrm{s}$ .

Step 2. Landing distance from the table's edge: $R = 2.0 \times 0.50 = \boxed{1.0\ \mathrm{m}}$ .

Step 3. The dropped marble lands at the foot of the table at the same $t = 0.50$ s — equal vertical histories. On the sketch: two curves leaving the same edge, one straight down, one a half-parabola, with rungs of equal height at equal times. The examiner is testing the belief, not the arithmetic: state "vertical motions are identical because horizontal velocity does not affect vertical acceleration" in words.

3 · Competitive level — the supply drop (JEE-Main/NEET pattern, ~2.5 minutes)

An aircraft flies horizontally at $100\ \mathrm{m/s}$ at a height of $500\ \mathrm{m}$ and releases a relief package ( $g = 10\ \mathrm{m/s^2}$ ). Find (a) the time for the package to land, (b) the horizontal distance it covers, (c) where the aircraft is relative to the package at the instant of landing.

Step 1. The package inherits the plane's velocity: horizontal launch at $u = 100\ \mathrm{m/s}$ from $h = 500$ m.

Step 2. Clock: $t = \sqrt{2(500)/10} = \sqrt{100} = \boxed{10\ \mathrm{s}}$ .

Step 3. Range: $R = 100 \times 10 = \boxed{1000\ \mathrm{m}}$ — release a full kilometre before the target.

Step 4. The plane keeps the same horizontal velocity as the package, so it is directly overhead at landing (500 m above the impact point). Landing speed, if asked: $v_y = 100\ \mathrm{m/s}$ , $v = \sqrt{100^2 + 100^2} = 100\sqrt{2} \approx 141\ \mathrm{m/s}$ at $45°$ below horizontal. Timing note: ~2.5 minutes, and parts (a)–(b) are 40 seconds of it. Part (c) is a concept checkpoint, not a calculation — "same $v_x$ , so directly above" is one sentence. If you find yourself computing the plane's position numerically, you've missed the physics the question is really buying.

4 · Competitive level — angled from the cliff, with a moat (JEE-Advanced stretch, ~4 minutes)

A ball is thrown from the top of a $20\ \mathrm{m}$ cliff at $25\ \mathrm{m/s}$ , angled $37°$ above the horizontal ( $\sin 37° = 0.6$ , $\cos 37° = 0.8$ , $g = 10\ \mathrm{m/s^2}$ ). A moat of width $70\ \mathrm{m}$ surrounds the cliff's base. (a) Find the flight time. (b) Does the ball clear the moat? (c) Find the landing speed.

Step 1. Components, down positive: $u_x = 20\ \mathrm{m/s}$ , $u_y = -15\ \mathrm{m/s}$ (upward start).

Step 2. Landing condition: $20 = -15t + 5t^2 \Rightarrow t^2 - 3t - 4 = 0 \Rightarrow (t-4)(t+1) = 0$ . Keep the honest root: $t = \boxed{4.0\ \mathrm{s}}$ .

Step 3. Range: $R = 20 \times 4 = 80\ \mathrm{m} > 70\ \mathrm{m}$ — clears the moat with 10 m to spare.

Step 4. Landing velocity: $v_x = 20$ ; $v_y = -15 + 10(4) = 25\ \mathrm{m/s}$ (down). Speed $v = \sqrt{400 + 625} = \sqrt{1025} \approx \boxed{32\ \mathrm{m/s}}$ . Energy cross-check: $v^2 = u_0^2 + 2gh = 625 + 400 = 1025$ ✓ — the launch angle never enters the landing speed, only its direction. Timing note: ~4 minutes, dominated by solving the quadratic cleanly. The factorable $(t-4)(t+1)$ is a setter's gift — if your quadratic refuses to factor, re-check the sign of $u_y$ before reaching for the formula. The energy shortcut in Step 4 turns part (c) into 15 seconds.

Exam traps

Putting $u$ into the vertical equation. The launch speed of a horizontal launch has zero vertical component. Writing $h = ut + \tfrac12gt^2$ manufactures a quadratic that shouldn't exist. Horizontal launch ⇒ vertical motion starts from rest. Full stop.

Using the level-ground range formula. $R = u_0^2\sin 2\theta/g$ from Projectile Motion assumes landing at launch height. From a cliff it fails — badly and silently. From a height, always rebuild from the two-step: vertical clock, then horizontal ruler.

Sign chaos in angled launches. Pick down positive, say so, and give the upward launch component its one minus sign. Mixing "up positive for the throw, down positive for the fall" mid-solution is the single most common source of impossible negative times.

Choosing the wrong root. The quadratic's negative root is not "the other answer" — it is a time before the throw. If both roots are positive, the surface was crossed twice (e.g. landing on a raised terrace); the question's geometry tells you which crossing is the landing.

Forgetting the inherited velocity. Objects released from moving vehicles carry the vehicle's velocity at the moment of release. A package "dropped" from a plane is not dropped; it is launched horizontally at the plane's speed.

Which $g$ ? CBSE/ICSE expect $g = 9.8\ \mathrm{m/s^2}$ unless stated; JEE-Main/Advanced and NEET nearly always declare $g = 10$ ; AP Physics 1 permits $10$ ; IB uses $9.8$ ; CIE A-Level uses $9.81$ . The clean $\sqrt{2h/g}$ numbers above are built on $g = 10$ — with $9.8$ expect $t = \sqrt{2(45)/9.8} = 3.03$ s, not 3 exactly. Use what the paper states, and never mix two values in one solution.

Landing "velocity" answered as a number only. Velocity is a vector; competitive papers award the direction. Give magnitude and angle below horizontal (or both components) unless the question says "speed".

Origins · 1 minIsaac Newton and how this was discovered

Isaac Newton

1642–1727 · Woolsthorpe & Cambridge → the Royal Mint

Newton took this archetype to its logical extreme. In A Treatise of the System of the World — the popular companion to the Principia — he drew a cannon on a mountaintop firing horizontally, harder and harder: each shot a projectile from a height, each range longer than the last. Fire hard enough, he argued, and the Earth's surface curves away exactly as fast as the ball falls toward it — the ball falls forever without landing. That single diagram (the "Newton's cannonball" figure, engraved around 1685) unified the table-edge marble with the Moon itself: orbital motion is free fall with enough horizontal speed. Your $t = \sqrt{2h/g}$ computation and a satellite's orbit are the same physics separated only by the size of $u$ — a story continued in Satellites: Speed, Period, Energy & Weightlessness.

History

1638

— Galileo's Two New Sciences proves the projectile path is a parabola by compounding uniform motion with free fall — using a ball rolling off a table's edge.

1685

— Newton's cannonball diagram extends the horizontal launch until falling becomes orbiting, unifying terrestrial and celestial mechanics.

1849

— Artillery ranging tables institutionalise the two-step method: height and gravity set the time, muzzle velocity sets the reach.

1940

s — Bombsight computers solve the supply-drop problem of Example 3 mechanically, in real time, under fire.

1957

— Sputnik realises the cannonball: a projectile from a height whose horizontal speed (about 7.8 km/s) never lets it land.

Going further

Texts. NCERT Physics Class 11, Ch. 4 (Motion in a Plane) · HC Verma, Concepts of Physics Vol. 1, Ch. 3 · Halliday–Resnick–Walker Ch. 4 (projectiles from elevated launch points) · Serway & Jewett, Ch. 4.

Provenance. All results are standard 2D kinematics. A verifier should recompute: $t = \sqrt{2h/g}$ and $R = u\sqrt{2h/g}$ from the two-step; Example 1's $t = 3$ s, $R = 60$ m, $v = \sqrt{1300} \approx 36$ m/s (45 m, 20 m/s, $g = 10$ ); Example 2's $t = 0.5$ s, $R = 1.0$ m; Example 3's $t = 10$ s, $R = 1000$ m, $v = 100\sqrt{2} \approx 141$ m/s; Example 4's quadratic $t^2 - 3t - 4 = 0 \Rightarrow t = 4$ s, $R = 80$ m, $v = \sqrt{1025} \approx 32$ m/s with the energy check $625 + 2(10)(20) = 1025$ ; the trajectory check $y(20) = 5$ m at $u = 20$ , $g = 10$ .

Prerequisites. Projectile Motion · Free Fall & Motion Under Gravity · Components & Resolution of Vectors. Leads to. Projectile on an Incline · Satellites: Speed, Period, Energy & Weightlessness.

Why does the launch speed not affect the time of flight?

Because time is decided entirely in the vertical direction, and a horizontal launch contributes nothing vertical. Gravity accelerates the projectile downward at $g$ whether it drifts sideways at 2 m/s or 200 m/s; the vertical equation $h = \tfrac12 gt^2$ simply has no $u$ in it. Independence of components is not an approximation or a trick — it is Newton's second law applied per axis, and the two-marbles demo is its experimental proof.

Why measure y downward? My textbook takes up as positive.

Both work; downward-positive just matches where the story happens. From a cliff, the displacement, the acceleration, and the final velocity all point down — measuring down-positive makes every term positive and the algebra unbreakable. Up-positive forces you to write $-h$ for the landing and $-g$ for the acceleration and keep them consistent for the whole solution. Choose once, declare it, never remix. For horizontal launches, down-positive removes literally every sign.

Where does the negative root of the angled-launch quadratic come from?

The quadratic describes the whole parabola, extended backward in time before your throw. The negative root is when a projectile flying this exact parabola would have passed ground level on the way up — a prequel that never happened, since yours began existing at $t = 0$ on the cliff-top. Keep the positive root. If a problem gives two positive roots, the trajectory really crosses the landing surface twice, and the geometry (a terrace, a wall) decides which crossing is meant.

Does the horizontal velocity really stay exactly constant the whole flight?

In the ideal (no air resistance) model, yes — nothing pushes horizontally, so $v_x = u$ from launch to landing. In reality drag bleeds horizontal speed, which is why long artillery shots and golf balls need corrections. Exams stay in the ideal model unless they explicitly introduce air resistance; if they do, expect only qualitative questions (shorter range, steeper descent), because the algebra leaves the algebra-and-trig world.

How is v² = u² + 2gh related to energy conservation?

It is energy conservation, wearing kinematic clothes. Multiply through by $\tfrac12 m$ : $\tfrac12 mv^2 = \tfrac12 mu^2 + mgh$ — final kinetic energy equals launch kinetic energy plus the potential energy surrendered in the fall. That is also why the landing speed in Example 4 ignores the launch angle: energy doesn't care about direction, only about the height dropped. The full story is the Work–Energy Theorem's opening act.

If the ball leaves the table horizontally, why does its path curve immediately instead of going straight for a while?

Gravity never waits. The instant support disappears, the acceleration is $g$ downward, and the parabola $y = gx^2/(2u^2)$ curves from the very first millimetre — it just curves gently at first because $v_y$ starts at zero. The cartoon picture of running off a cliff, hanging in the air, then falling is wrong precisely because it inserts a pause that no force can supply.

For the package-drop, why does the package stay directly below the plane? It looks like it falls behind in videos.

In the ideal model both keep the same horizontal velocity, so neither gains ground: the package stays on the plane's vertical line while dropping away from it. Videos show it drifting behind because air resistance retards the package (big drag, no engines) far more than the aircraft. The exam answer is "directly below, ignoring air resistance" — stating the ignoring earns the reasoning mark.