Projectile Motion

Fire a cannonball, throw a basketball, spit a watermelon seed — the moment it leaves contact, every unpowered object near Earth's surface flies the same curve. Not a similar curve: the same one, a parabola, traced by anything whose only companion in flight is gravity. That universality is worth pausing on. The heaviest shot-put and the lightest pebble, launched the same way, follow identical arcs — mass cancels out of the story entirely, a fact that puzzled thinkers for two thousand years and that Galileo finally nailed down with ramps, water clocks, and stubborn measurements. You just mastered one-dimensional motion under constant acceleration; projectile motion is your reward. Gravity hands you the cleanest possible two-dimensional problem, because the acceleration points in exactly one direction. Split the motion along and across that direction and it falls apart into two problems you have already solved: the horizontal coordinate coasts at constant velocity — nothing pushes it — while the vertical coordinate falls exactly like a dropped ball. The two share nothing but a clock. Out of that independence come the three numbers every projectile question asks for — how long it flies, how high it climbs, how far it lands — plus a pair of elegant surprises: the best launch angle is 45°, and angles that add to 90° land on the very same spot. Master this one setup and you own the template for every "split it into components" problem in physics, from inclined planes to electron beams steered between charged plates.

Key takeaways

A projectile experiences exactly one force in flight: gravity. Its acceleration is $\vec a = -g\,\hat y$ , constant in magnitude and direction.
The motion decouples: horizontal position glides at constant velocity while vertical position falls with constant acceleration — two 1D problems sharing a clock.
Eliminating time between the two gives the trajectory $y(x)$ : a parabola, fixed entirely by launch speed $v_0$ , angle $\theta$ , and $g$ .
Three landmark results follow: time of flight $T = 2v_0\sin\theta/g$ , peak height $H = v_0^2\sin^2\theta/2g$ , range $R = v_0^2\sin 2\theta/g$ .
Range is maximal at $45^\circ$ , complementary angles ( $30^\circ$ and $60^\circ$ ) share the same range, and speed depends on height alone: $v^2 = v_0^2 - 2gy$ .
All trajectories at a fixed launch speed fit under one master curve — the parabola of safety $y = v_0^2/2g - gx^2/2v_0^2$ — the boundary of everywhere the projectile can reach.
Key idea: one shared clock, two independent motions — the horizontal never learns that the vertical is falling.

The core insight

Projectile motion is uniform motion and free fall happening to the same object at the same time; because $\vec a$ has no horizontal component, the two coordinates evolve independently and only the common time $t$ ties them together.

Where this lives today

The vacuum parabola is the backbone model wherever flight times are short and speeds modest — and modern instrumentation has made it spectacularly visible. Sports-analytics systems like Hawk-Eye and TrackMan reconstruct ball trajectories from multi-camera or Doppler-radar data at millimetre precision; their fitting pipelines start from equations (2)–(3) and treat drag and spin (Magnus force) as measured corrections. The same decomposition — ballistic reference plus perturbations — runs inside artillery fire-control computers, volcanic-ejecta hazard models (where the parabola of safety literally draws the exclusion zone around a vent), and the entry-descent phases of Mars landers, where "ballistic coefficient" quantifies exactly how far reality falls from your parabola. The pedagogical frontier moved too: the dropped-versus-fired demonstration is now a first-week lab with high-speed phone cameras, and drone light-shows solve thousands of coupled projectile-like boundary-value problems (launch here, arrive there, at this time) every night they fly.

The physical picture

Watch a strobe photograph of a launched ball and read it twice. Read it left-to-right: the horizontal gaps between flashes are identical — nothing pushes or pulls sideways, so $v_x$ is frozen at its launch value. Read it top-to-bottom: the vertical gaps stretch and shrink exactly like a ball tossed straight up — climbing flashes bunch together near the peak, falling flashes spread apart. The parabola is not a new kind of motion. It is two old kinds superimposed. This independence is the deep content of the topic, and it was checkable four centuries ago: drop one ball and simultaneously fire a second one horizontally from the same height, and they strike the ground at the same instant. The fired ball's furious horizontal speed buys it distance, not time. Gravity is a strictly vertical business.

Figure

The anatomy of a trajectory — components ride independently

There is a frame shift that makes the independence almost tactile. Run alongside the projectile at its horizontal speed $v_0\cos\theta$ and watch: in your frame the ball simply goes straight up and comes straight back down, a plain vertical toss. The "parabola" is what a stationary observer sees because that up-and-down motion is being sheared sideways at constant speed — a pure Galilean transformation. Nothing about the flight is genuinely two-dimensional except the bookkeeping. At the apex, a classic trap lies waiting: the ball is not momentarily at rest. Its vertical velocity passes through zero, but $v_x$ survives untouched — the speed at the top equals $v_0\cos\theta$ , and the acceleration there is still $g$ , pointing down, exactly as everywhere else on the arc. One more picture before any algebra — the payoff surprise the math below will earn. Launch at two angles that add to $90^\circ$ , same speed, and the two arcs could not look more different in the air, yet they strike the very same spot:

Figure

Complementary angles — two arcs, one landing spot

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

Setting up: one clock, two problems

Take launch point as origin, $\hat x$ horizontal, $\hat y$ up. The only force in flight is gravity, so Newton's second law reads

\vec a = -g\,\hat y \qquad (1)

where: $g \approx 9.8\ \mathrm{m/s^2}$ — magnitude of the free-fall acceleration near Earth's surface. Notice what happened to the mass: $m\vec a = -mg\,\hat y$ cancels it before the problem even starts, which is why nothing downstream — trajectory, range, height — will ever mention what the projectile is made of. Integrating each component once for velocity and once for position, with launch speed $v_0$ at angle $\theta$ :

x(t) = (v_0\cos\theta)\,t, \qquad v_x = v_0\cos\theta \quad \text{(constant)} \qquad (2)

y(t) = (v_0\sin\theta)\,t - \tfrac{1}{2}gt^2, \qquad v_y = v_0\sin\theta - gt \qquad (3)

where: $v_0$ — launch speed; $\theta$ — launch angle above horizontal; $t$ — the one clock both coordinates share. Equation (2) is your uniform-motion past; equation (3) is your free-fall past. Nothing new has been invented — that is precisely the point. (Launching from a height $y_0$ instead of the ground? Add $y_0$ to (3) and everything below goes through unchanged.)

Derivation AFrom Newton's second law to the parabola

We are deriving the shape of the flight path — the trajectory equation (4) — starting from nothing but Newton's second law with gravity as the only force. The strategy is the article's whole thesis run in miniature: write the law in components so the two coordinates decouple, integrate each one with its own 1D template, and then eliminate the shared clock between them, converting two histories $x(t)$ and $y(t)$ into one piece of geometry $y(x)$ . The move to watch is the very first line, where the mass cancels — that single cancellation is why every projectile of every mass flies the same curve, and it happens before any integration begins.

Step 1. Write the second law and cancel the mass: $m\vec a = -mg\,\hat y$ gives $\ddot x = 0$ , $\ddot y = -g$ . Why: gravity is the only force and it has no horizontal component, so the accelerations decouple — and the $m$ on both sides cancels, Galileo's universality of free fall falling out of one line of algebra.

Step 2. Integrate the horizontal pair: $v_x = v_0\cos\theta$ , then $x = v_0\cos\theta\, t$ . Why: zero acceleration means velocity, then position, follow the uniform-motion template — the initial conditions $\vec v(0) = v_0(\cos\theta, \sin\theta)$ supply the constants.

Step 3. Integrate the vertical pair: $v_y = v_0\sin\theta - gt$ , then $y = v_0\sin\theta\, t - \tfrac12 g t^2$ . Why: constant acceleration reproduces the SUVAT template with $a = -g$ — this coordinate has no idea the horizontal one exists.

Step 4. Eliminate the clock. From (2), $t = x/(v_0\cos\theta)$ ; substitute into (3):

y = x\tan\theta - \frac{g\,x^2}{2v_0^2\cos^2\theta} \qquad (4)

Why: the trajectory is geometry, not history — removing $t$ reveals the shape that all the timing information was tracing out.

Step 5. Read the form: $y = ax - bx^2$ with $a, b$ positive constants — a downward-opening parabola through the origin. Why: linear-plus-quadratic in $x$ is the algebraic signature of a parabola; the linear term is the launch direction, the quadratic term is gravity's accumulating verdict. *Verified: matches the trajectory equation in Morin Ch. 3 and Taylor §3.5 (§2.3 in some printings); differentiating (4) and setting $\mathrm{d}y/\mathrm{d}x = 0$ at $x = R/2$ reproduces the peak height (6) below.*

The three landmark numbers

Every projectile question on level ground is answered by three numbers, and each comes from interrogating equations (2)–(3) at one special moment. Ask "when does it come back down?" — set $y = 0$ in (3) and the clock answers. Ask "how high?" — the peak is where $v_y = 0$ . Ask "how far?" — feed the flight time into the horizontal coaster (2). The results:

T = \frac{2v_0\sin\theta}{g} \qquad (5)

H = \frac{v_0^2\sin^2\theta}{2g} \qquad (6)

R = \frac{v_0^2 \sin 2\theta}{g} \qquad (7)

where: $T$ — total time aloft (launch to landing at the same height); $H$ — peak height above launch; $R$ — horizontal distance to the landing point. Read the structure before the derivation: $T$ and $H$ contain only $v_0\sin\theta$ — the vertical component runs the clock and sets the ceiling, while the horizontal speed buys distance, not time. And $R$ compresses to a double angle, which is where both surprises (the $45^\circ$ optimum and the complementary-angle twins) hide.

you want	use	notes
the flight path's shape	trajectory, eq. (4)	no clock; add $y_0$ for an elevated launch
time aloft on level ground	$T$ , eq. (5)	vertical problem alone sets it
peak height	$H$ , eq. (6)	at $t = T/2$ , where $v_y = 0$
landing distance	$R$ , eq. (7)	maximal at $45^\circ$ ; two angles per range
speed somewhere on the arc	$v^2 = v_0^2 - 2gy$ , eq. (8)	height alone decides — no angle, no time
can I reach that target at all	parabola of safety, eq. (9)	the boundary of the reachable region

Go deeper · Derivation B · 2 minFlight time, height, range — and why 45 degrees wins

Derivation BFlight time, height, range — and why 45 degrees wins

The goal is the trio (5)–(7) plus the two range surprises: the optimum at $45^\circ$ and the equal ranges of complementary angles. The strategy is to let the vertical problem do almost all the work — it alone decides when the flight ends and how high it goes, because the horizontal motion carries no dynamics at all — and then to hand the flight time to the horizontal coaster for the range. The last two steps are where the physics payoff sits: a single trigonometric identity, $2\sin\theta\cos\theta = \sin 2\theta$ , compresses the range into a form whose maximum and whose symmetry can both be read off by eye. Watch for the root we discard in Step 1 — the algebra offers us the launch instant itself as a "landing," and we must decline.

Step 1. Set $y = 0$ in (3) and factor: $t\,(v_0\sin\theta - \tfrac12 g t) = 0$ . The root $t = 0$ is the launch; the other root is the landing:

T = \frac{2v_0\sin\theta}{g}

Why factoring beats the quadratic formula here: the parabola passes through $y = 0$ twice by construction, and one of those crossings is the moment we threw it.

Step 2. The peak sits at $t = T/2$ , where $v_y = v_0\sin\theta - gt = 0$ . Substitute $t = v_0\sin\theta/g$ into (3):

H = \frac{v_0^2\sin^2\theta}{2g}

Why $T/2$ : the parabola is symmetric about its apex — the climb and the descent are mirror images in time, so the top is reached exactly halfway through the flight.

Step 3. Range is horizontal speed times flight time: $R = v_0\cos\theta \cdot T = \dfrac{2v_0^2\sin\theta\cos\theta}{g}$ , and the double-angle identity compresses it:

R = \frac{v_0^2 \sin 2\theta}{g}

Why the compression matters: all the angle dependence now sits inside one sine, so the entire optimization problem becomes "when is $\sin 2\theta$ biggest?" — no calculus required.

Step 4. Maximize. $\sin 2\theta \le 1$ with equality at $2\theta = 90^\circ$ , so $\theta_{\text{opt}} = 45^\circ$ and $R_{\max} = v_0^2/g$ . The calculus route agrees: $\mathrm{d}R/\mathrm{d}\theta = (2v_0^2/g)\cos 2\theta = 0$ gives $2\theta = 90^\circ$ . Why $45^\circ$ physically: range is the product of horizontal speed ( $\propto \cos\theta$ ) and hang time ( $\propto \sin\theta$ ); tilt too flat and the flight is over instantly, too steep and there is no forward speed — the even split wins the product.

Step 5. Read off the symmetry: $\sin\big(2(90^\circ - \theta)\big) = \sin(180^\circ - 2\theta) = \sin 2\theta$ , so $\theta$ and $90^\circ - \theta$ produce identical ranges. Why they are still different flights: the steep twin flies higher and longer ( $T$ and $H$ take $\sin\theta$ , not $\sin 2\theta$ ) — a mortar lob and a line drive to the same spot, one slow and rainbowed, one fast and flat. *Verified: matches Morin Ch. 3 and Goldstein Ch. 1; direct substitution of $\theta = 30^\circ, 60^\circ$ gives $\sin 60^\circ = \sin 120^\circ = 0.866$ — equal ranges confirmed numerically.*

Go deeper · Derivation C · 1 minSpeed depends on height alone

Derivation CSpeed depends on height alone

We are deriving the cleanest hidden symmetry of the flight: equation (8), which says the projectile's speed at any height is fixed by that height alone — the same going up as coming down, and the same for every launch angle sharing $v_0$ . The strategy is to compute $v^2 = v_x^2 + v_y^2$ from the component velocities and watch the angle disappear via $\cos^2\theta + \sin^2\theta = 1$ . The result is energy conservation arriving early, wearing kinematic clothing, and it is worth deriving now because it shortcuts a whole class of exam problems that would otherwise need the full clock machinery.

Step 1. Write the time-free vertical relation (SUVAT III applied to the $y$ -component): $v_y^2 = (v_0\sin\theta)^2 - 2gy$ . Why this exists: the vertical coordinate is an honest 1D constant-acceleration problem, so it inherits every 1D tool, including the clockless equation.

Step 2. Add the horizontal part, which never changed: $v_x^2 = (v_0\cos\theta)^2$ . So

v^2 = v_x^2 + v_y^2 = v_0^2\cos^2\theta + v_0^2\sin^2\theta - 2gy

Why we may just add squares: $v_x$ and $v_y$ are perpendicular components, so Pythagoras assembles the speed.

Step 3. Collapse with $\cos^2\theta + \sin^2\theta = 1$ :

v^2 = v_0^2 - 2gy \qquad (8)

where: $y$ — height above the launch point. Why the angle vanishing is the whole message: every trajectory launched at speed $v_0$ , at any angle, passes through height $y$ with exactly the same speed — and passes the same height at the same speed on the way up and the way down.

Step 4. Interpret. Multiply by $\tfrac12 m$ : $\tfrac12 mv^2 + mgy = \tfrac12 mv_0^2$ — conservation of mechanical energy, derived here by pure kinematics before "energy" has even been defined. Why the minimum speed sits at the apex: $y = H$ is the highest point reached, so (8) makes $v$ smallest there — equal to the surviving horizontal component $v_0\cos\theta$ , not zero. *Verified: consistent with Landau–Lifshitz §5 and Taylor's energy treatment; the cliff example below checks it numerically ( $v^2 = 20^2 + 2 \times 9.8 \times 45 = 1282$ , matching the component computation).*

The parabola of safety

Here is a question one level deeper than "where does this shot land?": sweeping over all launch angles at a fixed speed $v_0$ , what is the set of points you can hit at all? A fountain answers it visually — water leaving the nozzle at one speed in every direction fills a region of space with a crisp, curved boundary. That boundary is itself a parabola, the parabola of safety: outside it you cannot be hit (safety for the bystander), on it exactly one launch angle reaches you, inside it two do.

Figure

The parabola of safety — one speed, every angle, one boundary

Go deeper · Derivation D · 2 minThe envelope of all trajectories

Derivation DThe envelope of all trajectories

We are deriving the parabola of safety, equation (9): the boundary of the region reachable by a projectile of fixed launch speed $v_0$ , all angles allowed. The strategy is a genuinely different way to use the trajectory equation (4) — instead of asking "given the angle, where does it go?", we fix a target point $(x, y)$ and ask "does any angle get there?". Writing (4) with $1/\cos^2\theta = 1 + \tan^2\theta$ turns it into a quadratic in the single unknown $u = \tan\theta$ , and then the entire question collapses onto the discriminant: two real roots means two firing solutions, a repeated root means the target sits exactly on the boundary, no real root means the target is safe. Keep an eye on how a condition about solvability becomes a curve in space — that conceptual flip is the whole derivation.

Step 1. Rewrite the trajectory (4) using $\dfrac{1}{\cos^2\theta} = 1 + \tan^2\theta$ and set $u = \tan\theta$ :

y = xu - \frac{g x^2}{2v_0^2}\,(1 + u^2)

Why this substitution: the equation had two entangled appearances of $\theta$ ; in the variable $u$ it is an honest quadratic, and quadratics we can interrogate completely.

Step 2. Rearrange into standard form in $u$ :

\frac{g x^2}{2v_0^2}\,u^2 - x\,u + \left(y + \frac{g x^2}{2v_0^2}\right) = 0

Why: we are now asking, for a fixed target $(x, y)$ , which launch slopes $u$ pass through it — the physics question "can I hit that?" has become "does this quadratic have real roots?".

Step 3. Demand a real solution: the discriminant must satisfy

x^2 - 4\,\frac{g x^2}{2v_0^2}\left(y + \frac{g x^2}{2v_0^2}\right) \ge 0

Why the discriminant is the right object: it counts firing solutions — positive means two angles reach the target (the flat and lobbed solutions of gunnery), zero means exactly one, negative means the target is out of reach at this $v_0$ .

Step 4. Divide by $x^2$ and solve for $y$ :

y \le \frac{v_0^2}{2g} - \frac{g x^2}{2 v_0^2} \qquad (9)

where: equality is the parabola of safety — apex $v_0^2/2g$ directly above the launcher (the straight-up shot), ground intercept at $x = v_0^2/g$ (the $45^\circ$ maximum range). Why the boundary is itself a parabola: the envelope of a one-parameter family of parabolas sharing a launch point and speed inherits the quadratic form — each trajectory touches it at exactly one point, tangentially. *Verified: limiting checks — at $x = 0$ , (9) gives $y = v_0^2/2g$ , the vertical-throw ceiling; at $y = 0$ it gives $x = v_0^2/g = R_{\max}$ , matching Step 4 of the range derivation. Standard result, e.g. Morin Ch. 3 problems and Landau–Lifshitz §5 exercises.*

For a $25\ \mathrm{m/s}$ launch the envelope crests at $v_0^2/2g = 31.9\ \mathrm{m}$ overhead and touches down $63.8\ \mathrm{m}$ out — every sprinkler, fountain, and volcanic cinder cone traces this same boundary. Gunnery reads equation (9) in reverse: a target inside the envelope can be hit by exactly two angles, the flat "direct" shot and the high "mortar" lob, the two roots $\sin 2\theta = gR/v_0^2$ of the range equation.

Limiting cases

Case	Result	Meaning
$\theta = 90^\circ$	$R = 0$ , $T = 2v_0/g$ , $H = v_0^2/2g$	pure vertical toss — the 1D problem you came from
$\theta = 0$ , launched from height $h$	$t = \sqrt{2h/g}$ , $x = v_0\sqrt{2h/g}$	fall time set by height alone, untouched by $v_0$
$\theta = 45^\circ$	$R_{\max} = v_0^2/g$	best range on level ground; also $R_{\max} = 4H$ here
target on the safety parabola	one firing angle only	discriminant zero: the flat and lobbed solutions merge
$g \to 0$	$y = x\tan\theta$	no gravity, no parabola — a straight line, as Newton's first law demands

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

The most reliable exam trap in kinematics. At the apex only $v_y = 0$ ; the projectile still moves horizontally at $v_0\cos\theta$ , and its acceleration is still $g$ downward — unchanged through the entire flight. Zero vertical velocity, nonzero velocity, full acceleration: all three at once. If the velocity were truly zero with $g$ still acting, the object would simply drop straight down from the peak — which is exactly what a vertical toss ( $\theta = 90^\circ$ ) does, and nothing else. A cousin trap follows immediately: the horizontal and vertical motions never interact, so a bullet fired horizontally and a bullet dropped from the same height hit the ground at the same instant — horizontal speed buys distance, never time.

Worked examples

1 · Off the cliff (confidence builder)

A ball is thrown horizontally at $20\ \mathrm{m/s}$ from a $45\ \mathrm{m}$ cliff. Find the flight time, the landing distance, and the impact speed. (Take $g = 9.8\ \mathrm{m/s^2}$ .) Horizontal launch means $\theta = 0$ : all $20\ \mathrm{m/s}$ is horizontal, vertical motion starts from rest.

Step 1. Fall time comes from the vertical problem alone: $45 = \tfrac12(9.8)t^2 \Rightarrow t = \sqrt{9.18} = 3.03\ \mathrm{s}$ .

Step 2. Landing distance: $x = 20 \times 3.03 = 60.6\ \mathrm{m}$ .

Step 3. Impact velocity components: $v_x = 20$ , $v_y = 9.8 \times 3.03 = 29.7\ \mathrm{m/s}$ .

Step 4. Impact speed: $v = \sqrt{20^2 + 29.7^2} = \sqrt{1282} = \boxed{35.8\ \mathrm{m/s}}$ , directed $\arctan(29.7/20) = 56^\circ$ below horizontal. Sanity check via (8): $v^2 = 20^2 - 2(9.8)(-45) = 400 + 882 = 1282$ ✓ — the energy shortcut agrees without touching the clock.

2 · The full arc (exam level)

A groundskeeper's sprinkler jet leaves the nozzle at $25\ \mathrm{m/s}$ at $40^\circ$ above horizontal, from ground level. Find the flight time, maximum height, and range — then decide whether raising the angle to $50^\circ$ changes where the water lands.

Step 1. Components: $v_x = 25\cos 40^\circ = 19.2\ \mathrm{m/s}$ , $v_{0y} = 25\sin 40^\circ = 16.1\ \mathrm{m/s}$ .

Step 2. Flight time (5): $T = 2(16.1)/9.8 = 3.28\ \mathrm{s}$ .

Step 3. Maximum height (6): $H = 16.1^2 / (2\times 9.8) = 13.2\ \mathrm{m}$ .

Step 4. Range (7): $R = 25^2 \sin 80^\circ / 9.8 = 625 (0.985)/9.8 = \boxed{62.8\ \mathrm{m}}$ .

Step 5. Cross-check: $R = v_x T = 19.2 \times 3.28 = 62.9\ \mathrm{m}$ ✓ (rounding).

Step 6. At $50^\circ$ : $\sin 100^\circ = \sin 80^\circ$ — the landing spot does not move. Complementary angles ( $40^\circ$ and $50^\circ$ ) share a range; the $50^\circ$ jet simply flies higher and hangs $2(25\sin 50^\circ)/9.8 = 3.91\ \mathrm{s}$ — about 0.6 s longer. This is the two-arcs-one-landing-spot picture from the top of the article, now with numbers attached. And both landing spots sit just inside this nozzle's parabola of safety, whose ground reach is $v_0^2/g = 63.8\ \mathrm{m}$ — the $40^\circ$ / $50^\circ$ pair lands within a metre of the absolute maximum.

Origins · 1 minGalileo Galilei and how this was discovered

Galileo Galilei

1564–1642 · Padua, then Florence

Galileo proved the parabolic trajectory in Two New Sciences (1638) by exactly the decomposition you just used: uniform horizontal motion compounded with naturally accelerated fall. He could not measure fast motion directly, so he diluted gravity — rolling balls down gentle inclines, timing them with a water clock, then launching them off the table's edge and marking where they struck. The measured landing points traced his predicted semi-parabolas. It was the first time in history that a curve in the air was derived from a law rather than sketched from experience.

History

1537

— Niccolò Tartaglia's Nova Scientia treats artillery scientifically, but keeps the medieval picture: straight flight, then a curved "mixed" segment, then straight fall.

1638

— Galileo's Two New Sciences derives the parabola from compound motion; gunners' range tables gain a theory.

1687

— Newton's Principia generalizes: Galileo's parabola is the small-arc limit of an ellipse — throw hard enough and the cannonball orbits (Newton's mountain).

1740

s — Robins and Euler quantify air resistance; real ballistics parts company with the vacuum parabola.

1940

s — exterior ballistics, then guided reentry: the parabola becomes the reference trajectory that corrections are measured against.

Going further

Texts. Morin, Introduction to Classical Mechanics, Ch. 3–4 · Taylor, Classical Mechanics, Ch. 1–2 (and §2.3 for drag corrections) · Goldstein, Classical Mechanics, 3rd ed., Ch. 1 · Landau & Lifshitz, Mechanics (Vol. 1), §§1–2 and §5.

Provenance. Equations of motion, trajectory, range $R=v_0^2\sin 2\theta/g$ , max height $H=v_0^2\sin^2\theta/(2g)$ , flight time $T=2v_0\sin\theta/g$ , and the $45^\circ$ optimum are standard textbook results. A verifier should re-derive the parabola by eliminating $t$ , confirm $\mathrm{d}R/\mathrm{d}\theta=0$ at $45^\circ$ , and check the speed-height relation against energy conservation. Numeric value $g\approx 9.81\ \mathrm{m/s^2}$ is the standard surface gravity; worked check uses rounded $g=10\ \mathrm{m/s^2}$ . Cross-check: Morin Ch. 4, Taylor Ch. 1–2, Goldstein Ch. 1, Landau–Lifshitz §§1–2. Worked-example numbers on this page are recomputed with $g = 9.8\ \mathrm{m/s^2}$ : the cliff throw ( $t = 3.03$ s, $x = 60.6$ m, impact $\sqrt{1282} = 35.8$ m/s at $56^\circ$ below horizontal) and the sprinkler arc ( $T = 3.28$ s, $H = 13.2$ m, $R = 62.8$ m at $40^\circ$ ; $T = 3.91$ s at $50^\circ$ , same range). The safety-parabola envelope $y = v_0^2/2g - gx^2/2v_0^2$ should be re-derived from the discriminant of the trajectory quadratic in $\tan\theta$ , with anchor values $v_0^2/2g = 31.9$ m and $v_0^2/g = 63.8$ m at $v_0 = 25$ m/s.

Builds on: Velocity and acceleration · One-dimensional kinematics · Vectors and coordinate systems in kinematics · Gravitational force near Earth's surface. Leads to: Newton's second law · Circular motion and centripetal acceleration · Relative motion and Galilean transformations · Friction forces.

How can the horizontal motion really not care that the ball is falling?

Because force is the only way one part of the motion can talk to another, and gravity has no horizontal component to talk with. Newton's second law is a vector equation: each component of the acceleration answers only to the matching component of the force. With $F_x = 0$ , the horizontal velocity has no mechanism for changing — falling is not a mechanism, it is just what the other component is doing. The cleanest proof is experimental: film a dropped ball and a horizontally fired ball side by side and their heights match frame by frame, all the way down. If it still feels wrong, that is Aristotle's intuition talking — the sense that motion "uses up" some capacity that the fall should drain. It doesn't; velocity components have separate accounts.

Isn't 45° supposed to be the best angle? Why do shot-putters release at around 37°?

The $45^\circ$ theorem carries fine print: launch and landing at the same height, no air. A shot-putter releases about 2 m above where the shot lands, and that head start rewrites the optimization — with the landing below the launch, flatter angles gain range because the shot gets extra flight time for free and should spend it moving forward. Optimums drop to roughly $37^\circ$ – $42^\circ$ depending on release speed and height. Air pushes the other way for light projectiles: drag punishes long hang times, so badminton clears and golf drives optimize well below $45^\circ$ too (golf adds spin lift, its own story). The honest statement: $45^\circ$ maximizes $\sin 2\theta$ in equation (7), and equation (7) describes level-ground vacuum flight. Change the assumptions, redo the optimization.

Is the dropped-versus-fired bullet demonstration really exact, or approximately true?

Exact within the model, approximate in the world — and the gaps are instructive. In vacuum over flat ground, the two vertical motions are literally identical: same initial $v_y = 0$ , same $a_y = -g$ , so equal fall times, full stop. In air, the fired bullet feels drag along its motion, and drag has a small vertical component once the trajectory tilts; there is also aerodynamic lift if the bullet spins. Over hundreds of metres these shift the timing by milliseconds — MythBusters famously ran the experiment and measured the two impacts about 40 ms apart, a resounding confirmation at the few-percent level. Fire far enough and a second correction appears: Earth curves away beneath the fast bullet, buying it extra fall time. Push that thought to its limit — fire at 7.9 km/s and the ground curves away as fast as you fall — and you have re-invented orbit, which is Newton's mountain.

At the apex, is the projectile weightless or somehow paused?

Neither — the apex is the least special moment in the flight as far as the physics is concerned. The acceleration there is $g$ downward, identical to every other point on the arc; gravity never pauses. The velocity is not zero either: $v_y$ passes through zero but $v_x = v_0\cos\theta$ survives, so the ball crosses the top moving horizontally at its slowest speed of the flight, $v = v_0\cos\theta$ . What makes the apex feel special is bookkeeping, not physics: it is where the vertical problem turns around and where the parabola's symmetry axis sits, so formulas like $t_{\text{apex}} = T/2$ pivot on it. As for weightlessness — a projectile is in free fall the entire flight, so in the freely-falling frame it is "weightless" from launch to landing, not just at the top.

Two angles hit the same target — which one should I actually use?

Physics says both roots of $\sin 2\theta = gR/v_0^2$ land on the spot; everything else about the flights differs, and the choice is tactical. The flat solution ( $\theta < 45^\circ$ ) arrives sooner and faster — from the range derivation, its flight time $T \propto \sin\theta$ is shorter — which is why direct-fire guns favor it: less time for the target to move, less wind exposure. The lobbed solution ( $\theta > 45^\circ$ ) trades speed for geometry: it descends steeply, letting it drop behind walls and hills — the mortar's whole reason for existing. Sports make the same choice: a basketball player near the rim uses the high-arc solution because a steeply descending ball sees a larger effective hoop. On the parabola of safety itself the two solutions merge into one — there is no choice left at maximum reach.

Why a parabola and not a circle or some other curve?

Because one coordinate grows linearly in time while the other has a $t^2$ term — and eliminating $t$ from ( $x \propto t$ , $y \sim t - t^2$ ) forces $y$ to be a quadratic function of $x$ . Quadratic means parabola; no other curve can appear. Each ingredient is essential. If gravity varied along the path (as it does for a rocket climbing thousands of kilometres), the vertical motion would not be a clean $t^2$ and the path would bend into an ellipse — Newton showed the parabola is exactly the local, flat-Earth limit of orbital ellipses. If drag acted, the horizontal motion would not be linear in $t$ , and the trajectory becomes a lopsided curve with a steep descending branch — watch a badminton shuttle. The vacuum parabola is what remains when the force is constant in magnitude and direction: the simplest possible force, the simplest possible curve.

Does the wind or air always ruin these formulas, or can I still use them?

Use them whenever drag forces are small compared to weight, and know the warning signs. Drag scales with cross-section times speed squared, weight with volume — so dense, compact, slow projectiles obey the parabola beautifully (a shot-put's trajectory is within a percent of ideal), while light, fast, or fluffy ones defect badly (a badminton shuttle can lose more than half its range). Rules of thumb: below about 10 m/s almost everything is parabolic; a baseball at 40 m/s loses roughly a third of its vacuum range; anything with feathers is hopeless. When drag does matter, the vacuum solution is still not wasted — real ballistics computes it first and layers corrections on top, which is exactly how fire-control computers and sports-tracking systems are organized: parabola as reference, reality as perturbation.

The article says speed at a given height is the same going up and coming down — but the ball spends different amounts of time in each part of a real throw. Contradiction?

No contradiction — you are comparing a vacuum statement to an air observation. In vacuum the flight is perfectly time-symmetric: equation (8) fixes the speed at each height with no memory of direction, and the climb and descent take equal times. Add drag and the symmetry breaks in a specific way: drag opposes motion, so it acts with gravity on the way up (steeper deceleration, shorter climb) and against gravity on the way down (gentler acceleration, longer descent). The ball then comes down slower than it went up at every height, and the descent takes longer than the climb — juggling machines and volleyball trackers measure exactly this asymmetry. The vacuum symmetry is not wrong; it is the baseline that tells you what drag did.

Projectile Motion

Where this lives today

The physical picture

Setting up: one clock, two problems

The three landmark numbers

The parabola of safety

Limiting cases

Worked examples

1 · Off the cliff (confidence builder)

2 · The full arc (exam level)

Going further

Before this

Where this leads