Projectile Motion

Somewhere in every board paper, a ball is in flight. A footballer bends a free kick over the wall, a batter lofts a six over long-on, a stone leaves a catapult — and the question underneath is always the same one: the moment an object is on its own in the air, with nothing but gravity for company, what curve does it fly? The answer is startlingly democratic. A cricket ball and a crumpled wrapper, launched identically, trace the same parabola — mass never enters the story. Two thousand years of thinkers got this wrong before Galileo pinned it down with ramps, water clocks and a table edge. Here is the good news your last three modules have earned you: projectile motion contains nothing new. Gravity pulls straight down and only straight down, so the horizontal motion coasts at constant velocity while the vertical motion is exactly the free fall you have already mastered. Two old problems, running side by side, sharing nothing but a clock. Split the launch velocity into components, solve each axis with equations you already own, and every projectile question in CBSE, JEE-Main, NEET or AP falls open. Exams love this topic beyond all proportion — it is the most-asked 2D setup in kinematics — because it tests everything at once: components, suvat, symmetry, nerve. Master the three landmark formulas and the trajectory equation here, and the archetypes that follow (towers, inclines) become variations on a theme you already hum.

Key takeaways

In flight, a projectile feels exactly one force: gravity. So $a_x = 0$ and $a_y = -g$ — the horizontal motion is uniform, the vertical motion is free fall, and the two share only the time $t$ .
Resolve the launch velocity $u$ at angle $\theta$ : $u_x = u\cos\theta$ stays frozen for the whole flight; $u_y = u\sin\theta$ dies, reverses and returns by symmetry.
Three landmark results on level ground: time of flight $T = \dfrac{2u\sin\theta}{g}$ , maximum height $H = \dfrac{u^2\sin^2\theta}{2g}$ , range $R = \dfrac{u^2\sin 2\theta}{g}$ .
Range peaks at $\theta = 45^\circ$ , and complementary angles ( $\theta$ and $90^\circ - \theta$ ) land on the same spot with different flights.
Eliminating $t$ gives the trajectory $y = x\tan\theta - \dfrac{gx^2}{2u^2\cos^2\theta} = x\tan\theta\left(1 - \dfrac{x}{R}\right)$ — a parabola, and a JEE favourite in its second form.
Key idea: one shared clock, two independent motions — the horizontal part never finds out that the vertical part is falling.

The core insight

Projectile motion is uniform horizontal motion and vertical free fall happening to the same object at the same time; because gravity has no horizontal component, the two coordinates evolve independently, tied together only by the common clock $t$ — so every formula of the topic is a 1D result you already know, dressed in components.

Where this lives today

Sports analytics runs on this module. Hawk-Eye's cameras and TrackMan's Doppler radar reconstruct ball flight in cricket, tennis, golf and baseball by fitting the measured arc against equations (2) and (3), with drag and spin as corrections on top; basketball shot-tracking reports each player's launch angle because the flat shot and the high rainbow are complementary-angle cousins with very different rim tolerance. Artillery fire-control computers still begin from the vacuum parabola before layering in air density and wind — the fire tables Tartaglia dreamed of, finally correct. And every ornamental fountain is frozen trajectory design: nozzle speed and angle chosen so the water's parabola lands exactly in the catch basin. The dropped-versus-fired demonstration is now a first-week lab with a phone's slow-motion camera — two coins, one table edge, one simultaneous thud.

The physical picture

Watch a strobe photograph of a launched ball and read it twice. Left to right: the horizontal gaps between flashes are identical — nothing pushes or pulls sideways, so $u\cos\theta$ survives from launch to landing without losing a single m/s. Top to bottom: the vertical gaps bunch together near the peak and stretch apart on the way down, exactly like a ball tossed straight up. The parabola is not a new kind of motion. It is two old kinds superimposed — and the axis-independence principle you met in Motion in a Plane is doing all the work.

Figure

The anatomy of a trajectory — the two components ride independently

The independence was checkable four centuries ago and still stuns students today: drop one ball and simultaneously fire a second one horizontally from the same height. They hit the ground at the same instant, every time. The fired ball's furious horizontal speed buys it distance — never time. Gravity is a strictly vertical business. One more picture before the algebra, because it is the payoff surprise the math below will earn. Launch at two angles that add to $90^\circ$ , same speed: the arcs could not look more different in the air, yet they strike the very same spot.

Figure

Complementary angles — two very different arcs, one landing spot

The flat $35^\circ$ line drive and the towering $55^\circ$ rainbow are range-twins: what the steep shot loses in horizontal speed it gains, in exactly even trade, in hang time. Why the trade is perfectly even — and why $45^\circ$ splits the difference for maximum range — falls out of one trigonometric identity below.

One clock, two problems

Set the launch point as origin, $x$ horizontal, $y$ vertically up, and resolve the launch velocity the way Components and Resolution of Vectors taught you. HS boards write initial speed as $u$ , so the components are $u_x = u\cos\theta$ and $u_y = u\sin\theta$ . The only in-flight force is gravity, so the acceleration is the same at every point of the arc:

a_x = 0, \qquad a_y = -g \qquad (1)

where: $g$ is the free-fall acceleration ( $9.8\ \mathrm{m/s^2}$ ; board numericals usually say "take $g = 10\ \mathrm{m/s^2}$ "). The minus sign encodes our up-is-positive convention. Equation (1) is the entire physics of the topic — everything after it is bookkeeping. Feed (1) into the kinematic equations per axis. The horizontal axis, with zero acceleration, keeps its uniform-motion form:

x = (u\cos\theta)\,t, \qquad v_x = u\cos\theta \ \ \text{(never changes)} \qquad (2)

where: $t$ is the shared clock started at launch. Read (2) physically: the horizontal coordinate has no idea a flight is happening — it is a walker pacing steadily along the ground, and at the end of the flight it will have paced $u\cos\theta \times T$ . The vertical axis is a ball thrown straight up at speed $u\sin\theta$ — your Free Fall module verbatim:

y = (u\sin\theta)\,t - \tfrac{1}{2}gt^2, \qquad v_y = u\sin\theta - gt \qquad (3)

where: $v_y$ falls through zero at the apex and returns to $-u\sin\theta$ at landing — the up-down symmetry of free fall, inherited whole. Nothing in (2) or (3) is new; that is precisely the point of the topic.

you are asked for…	work in…	using
time (of flight, to a height)	vertical axis only	equation (3)
horizontal distance	horizontal axis, after the clock is known	equation (2)
speed or velocity at a point	both components, then combine	$v = \sqrt{v_x^2 + v_y^2}$
shape of the path / "y when x = …"	eliminate $t$	trajectory equation (7)

Derivation ATime of flight and maximum height from the vertical clock alone

We want $T$ , the total airborne time on level ground, and $H$ , the peak height — and we want to see that both come entirely from the vertical problem, with the horizontal motion contributing nothing. The strategy: apply the 1D equations along $y$ with $a = -g$ , use the apex condition $v_y = 0$ , and lean on up-down symmetry. Watch that it is $u\sin\theta$ , not $u$ , that plays the role of "launch speed" here — mixing those up is the most common substitution slip in this chapter.

Step 1. At the apex the vertical velocity is zero. Set $v_y = 0$ in (3): $0 = u\sin\theta - gt_{\text{up}}$ , so the climb takes $t_{\text{up}} = \dfrac{u\sin\theta}{g}$ . Why only the vertical: the apex is defined by the ball momentarily not rising or falling — the horizontal component has no say in when that happens.

Step 2. By free-fall symmetry the descent to launch height retraces the climb in equal time, so the total flight is

T = \frac{2u\sin\theta}{g}

Step 3. For the height, use the timeless equation $v^2 = u^2 + 2as$ along $y$ with $v_y = 0$ at the top: $0 = (u\sin\theta)^2 - 2gH$ , giving

H = \frac{u^2\sin^2\theta}{2g}

Why the timeless route: we neither know nor need the time — height and vertical speed are linked directly, which is exactly what that equation is for.

Step 4. Read the pair. Both $T$ and $H$ contain only $u\sin\theta$ : two projectiles with equal vertical launch components stay up equally long and climb equally high, however different their horizontal speeds. Hang time cannot be bought with forward speed. Verified: for u = 20 m/s, θ = 30°, g = 10 m/s²: u sin θ = 10 m/s, so T = 2(10)/10 = 2.0 s and H = 10²/(2×10) = 5.0 m; the apex sits at t = 1.0 s, half of T, as symmetry demands.

Go deeper · Derivation B · 1 minRange and the double angle that hides inside it

Derivation BRange and the double angle that hides inside it

We want the range $R$ — the horizontal distance covered when the projectile returns to launch height — and the two famous facts it carries: maximum at $45^\circ$ , and equal ranges for complementary angles. The strategy is one honest multiplication: range is horizontal speed times time aloft, because the horizontal motion is uniform. The only subtlety is a trigonometric identity that compresses the answer and then hands us both symmetry results for free.

Step 1. The walker of equation (2) paces at $u\cos\theta$ for the full flight time $T$ from the previous derivation:

R = (u\cos\theta)\,T = (u\cos\theta)\,\frac{2u\sin\theta}{g} = \frac{2u^2\sin\theta\cos\theta}{g}

Step 2. Compress with the double-angle identity $2\sin\theta\cos\theta = \sin 2\theta$ :

R = \frac{u^2\sin 2\theta}{g}

Why bother compressing: the physics of the whole angle-dependence now sits in one factor, $\sin 2\theta$ , and everything we want to know about "which angle is best" becomes a question about the sine function.

Step 3. Maximum range: $\sin 2\theta$ peaks at $1$ when $2\theta = 90^\circ$ , i.e. $\theta = 45^\circ$ , giving $R_{\max} = \dfrac{u^2}{g}$ . The best launch splits the velocity evenly between buying time and buying speed.

Step 4. Complementary angles: replacing $\theta$ by $90^\circ - \theta$ turns $2\theta$ into $180^\circ - 2\theta$ , and $\sin(180^\circ - 2\theta) = \sin 2\theta$ . So $\theta$ and $90^\circ - \theta$ give identical ranges — with different times of flight and heights. The steep twin trades horizontal speed for hang time at a perfectly even exchange rate. Verified: u = 20 m/s, θ = 30°, g = 10 m/s² gives R = 400 × sin 60° / 10 = 400 × 0.866 / 10 = 34.6 m; cross-check by the definition, R = u cos θ × T = 17.32 × 2.0 = 34.6 m; the complementary launch at 60° gives R = 400 × sin 120° / 10 = 34.6 m again, while its T = 3.46 s and H = 15 m are entirely different.

Go deeper · Derivation C · 1 minThe trajectory equation — eliminating the clock

Derivation CThe trajectory equation — eliminating the clock

We want the shape of the path, $y$ as a function of $x$ , with time removed — the equation that answers "will it clear the wall at $x = 30$ m?" in one substitution. The strategy: solve the horizontal equation (2) for $t$ , substitute into the vertical equation (3), and then massage the result into the exam-friendly form built from $\tan\theta$ and $R$ . Keep an eye on where $\cos^2\theta$ lands — it is the piece students misplace.

Step 1. From (2), the clock reads $t = \dfrac{x}{u\cos\theta}$ when the projectile is at horizontal distance $x$ .

Step 2. Substitute into (3):

y = (u\sin\theta)\frac{x}{u\cos\theta} - \frac{1}{2}g\left(\frac{x}{u\cos\theta}\right)^2 = x\tan\theta - \frac{g\,x^2}{2u^2\cos^2\theta}

Why this is a parabola: y is a linear term minus a quadratic term in x with constant coefficients — the algebraic signature of a downward-opening parabola through the origin.

Step 3. Read the two terms physically. The first, $x\tan\theta$ , is the straight aim line — where the ball would be if gravity were switched off. The second is gravity's theft: exactly $\tfrac12 gt^2$ , the free-fall drop, rewritten in terms of $x$ . A projectile is always found directly below its no-gravity self.

Step 4. For the JEE-favourite form, factor out $x\tan\theta$ and use $R = \dfrac{2u^2\sin\theta\cos\theta}{g}$ to recognise the coefficient:

y = x\tan\theta\left(1 - \frac{x}{R}\right)

Why examiners love this form: it needs only the angle and the range — no u, no g — and it makes the endpoints obvious: y = 0 at x = 0 and at x = R. *Verified: at half range x = R/2 the form gives y = (R/2)tan θ × (1/2) = R tan θ/4; substituting R = u² sin 2θ/g reduces this to u² sin²θ/2g = H exactly — the parabola's midpoint is its peak. Numbers: u = 20, θ = 30°, g = 10: at x = 17.3 m, y = 17.3 × 0.577 × 0.5 = 5.0 m = H.*

Limiting cases

case	result	meaning
$\theta = 90^\circ$	$R = 0$ , $T = 2u/g$ , $H = u^2/2g$	a pure vertical throw — this is exactly your Free Fall module
$\theta = 0^\circ$ (level ground)	$T = 0$ , $R = 0$	launched along the ground it lands immediately; height launches belong to the tower archetype
$\theta = 45^\circ$	$R_{\max} = u^2/g$ , and $R = 4H$	best range on level ground; height is a quarter of the range
$\theta$ and $90^\circ - \theta$	same $R$ , different $T$ , $H$	the complementary twins — $\sin 2\theta$ cannot tell them apart
$g \to$ larger (other planets)	$T$ , $H$ , $R$ all shrink as $1/g$	same throw on the Moon ( $g/6$ ) flies six times farther
$u \to 2u$	$T \times 2$ , $H \times 4$ , $R \times 4$	$T$ is linear in $u$ ; $H$ and $R$ are quadratic — scaling MCQs live here

Common trap“At the top of the flight, the velocity is zero”

The single most reliable mark-killer in this chapter. At the apex only the vertical component $v_y$ is zero. The projectile is still moving — horizontally, at the untouched $u\cos\theta$ — so its speed at the top is $u\cos\theta$ , its smallest but never zero (unless $\theta = 90^\circ$ , a vertical throw). And the acceleration at the apex is still $g$ , downward, exactly as everywhere else on the arc: gravity does not pause because the climb ended. Test the claim yourself: if the velocity were truly zero with $g$ still acting, the ball would drop straight down from the peak — only a vertically-thrown ball does that. MCQ setters plant "v = 0, a = 0 at the highest point" every single year; both halves are wrong.

Worked examples

Board level — the full flight, marks-scheme style (CBSE/ICSE pattern)

A football is kicked from ground level with speed $20\ \mathrm{m/s}$ at $30^\circ$ above the horizontal. Take $g = 10\ \mathrm{m/s^2}$ . Find (a) the time of flight, (b) the maximum height, and (c) the horizontal range. (d) State the speed of the ball at its highest point.

Step 1. Resolve the launch velocity — the step the marks scheme pays for first: $u_x = 20\cos 30^\circ = 17.32\ \mathrm{m/s}$ , $u_y = 20\sin 30^\circ = 10\ \mathrm{m/s}$ .

Step 2. Time of flight from the vertical problem: $T = \dfrac{2u\sin\theta}{g} = \dfrac{2 \times 10}{10} = \boxed{2.0\ \mathrm{s}}$ .

Step 3. Maximum height: $H = \dfrac{u^2\sin^2\theta}{2g} = \dfrac{400 \times 0.25}{20} = \boxed{5.0\ \mathrm{m}}$ .

Step 4. Range: $R = \dfrac{u^2\sin 2\theta}{g} = \dfrac{400 \times \sin 60^\circ}{10} = \dfrac{400 \times 0.866}{10} = \boxed{34.6\ \mathrm{m}}$ . Cross-check with the definition: $R = u_x T = 17.32 \times 2.0 = 34.6\ \mathrm{m}$ ✓.

Step 5. At the highest point $v_y = 0$ , so the speed is the surviving horizontal component: $\boxed{17.32\ \mathrm{m/s}}$ — not zero. Writing "0" here is the classic lost mark.

Competitive level — same range, two flights (JEE-Main pattern)

A projectile launched at $40\ \mathrm{m/s}$ has the same range for two angles of projection. The shorter time of flight is $4\ \mathrm{s}$ . Take $g = 10\ \mathrm{m/s^2}$ . Find the longer time of flight and the common range.

Step 1. Same range at fixed $u$ means complementary angles $\theta$ and $90^\circ - \theta$ . Their times of flight are $t_1 = \dfrac{2u\sin\theta}{g}$ and $t_2 = \dfrac{2u\cos\theta}{g}$ .

Step 2. Use $t_1 = 4$ s: $\sin\theta = \dfrac{g t_1}{2u} = \dfrac{10 \times 4}{80} = 0.5$ , so $\theta = 30^\circ$ (the flatter twin owns the shorter flight).

Step 3. Longer flight: $t_2 = \dfrac{2 \times 40 \times \cos 30^\circ}{10} = 8 \times 0.866 = \boxed{6.93\ \mathrm{s}}$ (that is $4\sqrt{3}$ s).

Step 4. Range: $R = \dfrac{u^2 \sin 60^\circ}{g} = \dfrac{1600 \times 0.866}{10} = \boxed{138.6\ \mathrm{m}}$ .

Step 5. Bank the general identity this problem is built on: multiply the two flight times — $t_1 t_2 = \dfrac{4u^2\sin\theta\cos\theta}{g^2} = \dfrac{2R}{g}$ , so $R = \tfrac{1}{2}g\,t_1 t_2$ . Check: $\tfrac12 \times 10 \times 4 \times 6.93 = 138.6$ m ✓. Many papers ask for $R$ directly from $t_1$ and $t_2$ — this one line answers them. Timing note: 90 seconds with $R = \tfrac12 g\,t_1t_2$ memorised; 3 minutes rebuilding from components. The identity is worth banking — it recurs in JEE-Main archives every few years.

Competitive level — when height equals range (NEET pattern)

For what angle of projection is the maximum height equal to the horizontal range?

Step 1. Set $H = R$ : $\dfrac{u^2\sin^2\theta}{2g} = \dfrac{2u^2\sin\theta\cos\theta}{g}$ — use the $2\sin\theta\cos\theta$ form of the range so both sides share factors.

Step 2. Cancel $u^2/g$ and one $\sin\theta$ (nonzero for a genuine launch): $\dfrac{\sin\theta}{2} = 2\cos\theta$ .

Step 3. Rearrange: $\tan\theta = 4$ , so $\theta = \tan^{-1}4 \approx \boxed{76^\circ}$ .

Step 4. Sanity check against the general ratio $\dfrac{H}{R} = \dfrac{\tan\theta}{4}$ (divide the two formulas): $H = R$ forces $\tan\theta = 4$ ✓. This little ratio is another one-liner worth banking — it converts every " $H$ is such-and-such fraction of $R$ " question into a single tangent. Timing note: 40 seconds with $H/R = \tan\theta/4$ banked; the option grid will offer $\tan^{-1}2$ and $45^\circ$ as decoys for students who dropped a factor of 2.

Exam traps

"Velocity is zero at the top." Only $v_y$ is. Speed at the apex is $u\cos\theta$ ; acceleration there is $g$ , unchanged. Both the "v = 0" and the "a = 0" versions appear as MCQ statements — reject both.

$\sin 2\theta$ butchered. It is $\sin(2\theta)$ , not $2\sin\theta$ — for $\theta = 30^\circ$ that is $\sin 60^\circ = 0.866$ , not $2 \times 0.5 = 1$ . And keep the calculator in degree mode; a radian-mode $\sin 60$ silently wrecks the paper.

Range formula used off a cliff. $R = u^2\sin 2\theta/g$ is valid only when landing height equals launch height. Tower, cliff and balcony launches need the full vertical equation — that setup has its own archetype module (Projectile from a Height).

Which g? CBSE, JEE and NEET numericals almost always say "take $g = 10\ \mathrm{m/s^2}$ " — use it, or your options will not match. AP, IB and A-Level work with $9.8$ ( $9.81$ on A-Level data sheets). Read the paper's data line before substituting.

Angle measured from the vertical. Some setters give " $30^\circ$ with the vertical" — that is $60^\circ$ with the horizontal, and every sin and cos swaps. Underline the reference line in the question before resolving.

Complementary decoys. "Same range" never means "same flight": the steeper twin flies higher and longer ( $t_1 t_2 = 2R/g$ ties them). Options asserting equal times or equal heights for $30^\circ$ / $60^\circ$ are planted traps.

Forgetting that $v_x$ never changes. On level ground the landing speed equals the launch speed $u$ (symmetry), and the landing angle equals the launch angle below horizontal. Options offering a smaller landing speed test exactly this.

Origins · 1 minGalileo Galilei and how this was discovered

Galileo Galilei

1564–1642 · Padua, then Florence

Galileo owned the parabola. In the Fourth Day of his Two New Sciences (1638) he proved that a projectile's path is a parabola by compounding two motions he had mastered separately — uniform horizontal motion and naturally accelerated fall — insisting that neither disturbs the other. Unable to time fast flights directly, he diluted gravity: bronze balls rolled down gentle inclines, shot off the table edge, landing points marked for each launch speed. The measured points traced his predicted semi-parabolas — the first time a curve through the air was derived from a law rather than sketched from a gunner's experience. His student Evangelista Torricelli extended the result into range tables and the envelope of all possible shots. The independence argument you used on the strobe photograph is Galileo's, unchanged after four centuries.

History

1537

— Niccolò Tartaglia's Nova Scientia brings mathematics to artillery but keeps the medieval picture: straight flight, a curved middle, then straight fall.

1638

— Galileo's Two New Sciences, Fourth Day: compound motion proves the trajectory is a parabola.

1644

— Torricelli works out the envelope of all trajectories — the safety parabola no shot can cross — and practical range tables.

1687

— Newton's Principia stretches the idea to its limit: thrown hard enough, the cannonball's parabola bends into an orbit (Newton's mountain).

2001

— Hawk-Eye multi-camera tracking debuts in cricket broadcasting; the vacuum parabola becomes the reference curve inside live sports graphics.

Going further

Texts. NCERT Physics XI Part 1, Ch. 4 (Motion in a Plane) — the CBSE/NEET canonical treatment · H.C. Verma, Concepts of Physics Vol. 1, Ch. 3 · Halliday, Resnick & Walker, Ch. 4 · Galileo, Two New Sciences, Fourth Day — the original parabola proof. The full vector-calculus treatment with drag is the UG deep dive Projectile Motion (aq-mech tier).

Provenance. All results are standard kinematics per NCERT Ch. 4. A verifier should recompute: the anchor set $u = 20$ m/s, $\theta = 30^\circ$ , $g = 10$ : $T = 2.0$ s, $H = 5.0$ m, $R = 34.6$ m $= u_xT = 17.32 \times 2.0$ ; the trajectory midpoint check $y(R/2) = R\tan\theta/4 = H$ ; the complementary pair at $60^\circ$ : $T = 3.46$ s, $H = 15$ m, same $R$ ; JEE example $\sin\theta = 40/80 = 0.5$ , $t_2 = 4\sqrt3 = 6.93$ s, $R = 1600 \times 0.866/10 = 138.6 = \tfrac12(10)(4)(6.93)$ ; NEET example $H = R \Rightarrow \tan\theta = 4$ , $\theta = 76^\circ$ ; scaling row $u \to 2u$ : $T \times 2$ , $H, R \times 4$ .

Prerequisites. Motion in a Plane — the axis-independence principle this module applies · Free Fall — the vertical half of every flight. Leads to. Projectile from a Height — towers, cliffs and the full quadratic · Projectile on an Incline — range along a slope.

Why is 45° the best angle, and why do 30° and 60° give the same range?

Range is (horizontal speed) × (time aloft). Aim low and you pace fast but land almost immediately; aim high and you hang forever while barely drifting forward. The identity $2\sin\theta\cos\theta = \sin 2\theta$ puts the whole compromise into one factor, maximised when $2\theta = 90^\circ$ — $45^\circ$ , the perfectly even split. For $30^\circ$ and $60^\circ$ the doubles are $60^\circ$ and $120^\circ$ , whose sines are equal: the steep shot's extra hang time exactly compensates its slower pacing. Equal range — but not equal flight: the $60^\circ$ shot takes $\sqrt{3}$ times longer.

Does the horizontal velocity really stay constant? What about air resistance?

In the exam model, yes, exactly: no horizontal force means no horizontal acceleration, so $v_x = u\cos\theta$ from launch to landing. In real air, drag opposes motion and does bleed horizontal speed — which is why a real cricket ball's arc is slightly steeper coming down than going up, and why real maximum-range angles are below $45^\circ$ (a javelin's is nearer $35^\circ$ ). Every board and entrance exam states or assumes "neglect air resistance" for this topic; use the ideal model unless the paper explicitly says otherwise, and mention air resistance only when asked why real trajectories differ.

When should I use the trajectory equation instead of the suvat components?

Use the trajectory equation when the question is about geometry with time absent: will it clear a wall of height $y$ at distance $x$ , what is $y$ when $x$ is so-and-so, where does the parabola sit relative to a target. Use the component equations when time is in the story: how long, when is it at height $h$ , what is the velocity after 2 s. If a JEE question hands you the angle and the range only, the $y = x\tan\theta(1 - x/R)$ form is usually a one-line finish — that is exactly why it is worth memorising alongside the standard form.

Why does the time of flight depend only on the vertical component?

Because "in the air" is a vertical condition — the flight ends when the ball comes back down to launch height, and coming down is governed entirely by the vertical motion. The horizontal component moves the ball along, never up, so it has no vote on when the flight ends. Convince yourself with the two-ball experiment: the dropped ball and the horizontally-fired ball hit the ground together because their vertical problems are identical, even though one of them is also travelling sideways at high speed. Forward speed buys distance, never time.

Why doesn't mass appear in any projectile formula?

Gravity's force on a projectile is $mg$ , and by Newton's second law its acceleration is $mg/m = g$ — the mass cancels before the motion starts. Every formula here is built from $g$ , $u$ and $\theta$ alone: a shot-put and a tennis ball launched identically trace identical arcs in vacuum. In real air, mass sneaks back: drag depends on size and speed but not mass, so it decelerates a light object more than a heavy one of the same shape — why a paper ball and a cricket ball part company in practice while agreeing in theory.

The answer key says the landing speed equals the launch speed. How can that be, after the whole flight?

On level ground it is exact, by symmetry. The horizontal component never changes, and the vertical component returns with the magnitude it left with ( $u\sin\theta$ , now downward) by the up-down symmetry of free fall. Same component magnitudes, same combined speed: $u$ . The landing velocity is different — it points below the horizontal at angle $\theta$ — exactly the speed-versus-velocity distinction exams probe. If the landing point is higher or lower than the launch, the symmetry breaks and the landing speed changes; that is tower-archetype territory.

What changes if the projectile is launched from a height, or on a slope?

The physics — equations (1) to (3) — changes not at all; only the landing condition does. From a tower the flight ends at $y = -h$ , so you solve the full quadratic in $t$ instead of using $T = 2u\sin\theta/g$ , and the neat $R$ and symmetry results stop applying. On an incline, "landing" means meeting the slope line $y = x\tan\alpha$ . Both setups are drilled in their own archetype modules (Projectile from a Height, Projectile on an Incline); misapplying $R = u^2\sin 2\theta/g$ off a cliff is precisely the trap they open with.