Projectile Motion — Fundamentals

In short

A projectile launched at speed u and angle \theta from the ground follows a parabolic path. Its horizontal position is x = u\cos\theta \cdot t (constant speed, no horizontal force) and its vertical position is y = u\sin\theta \cdot t - \tfrac{1}{2}g t^2 (gravity pulls it down). The time of flight is T = \frac{2u\sin\theta}{g}, the maximum height is H = \frac{u^2\sin^2\theta}{2g}, and the horizontal range is R = \frac{u^2\sin 2\theta}{g}. The range is maximum at \theta = 45°, and complementary angles (\theta and 90° - \theta) give the same range.

A Diwali rocket shoots up from the street at an angle, traces a bright arc across the dark sky, and falls back to the ground some distance away. A cricket batsman swings hard and sends the ball soaring over the bowler's head — it rises, hangs in the air for a moment, then curves down into the stands for a six. A water jet from a garden hose, aimed upward at an angle, traces a graceful curve before splashing onto the lawn.

All three are projectiles. The moment the object leaves the launcher — the rocket tube, the bat face, the hose nozzle — it has no engine, no thrust, no horizontal force. The only force acting on it is gravity, pulling straight down. And yet the object does not fall straight down. It follows a curve. Understanding that curve — its shape, its height, its length — is what projectile motion is about.

Two motions, completely independent

Here is the key insight that makes projectile motion simple: a projectile's horizontal and vertical motions are independent of each other. Gravity acts only downward. It has no horizontal component. So whatever the object is doing horizontally, gravity does not interfere with it. And whatever gravity is doing vertically, the horizontal motion does not interfere with that either.

Think of it this way. Imagine you are standing on the roof of a building. You hold two cricket balls at the same height. At the exact same instant, you drop one straight down and throw the other one horizontally. Which ball hits the ground first?

Your intuition might say the dropped ball — after all, the thrown ball has to travel a longer path. But that intuition is wrong. Both balls hit the ground at the same instant. The thrown ball moves sideways as it falls, but its vertical fall is identical to the dropped ball's. Gravity does not care whether the ball is also moving horizontally. It pulls both balls down at the same rate: 9.8 \text{ m/s}^2.

The initial velocity $u$ at angle $\theta$ decomposes into a horizontal component $u\cos\theta$ (constant, because no horizontal force acts) and a vertical component $u\sin\theta$ (which gravity decelerates at $9.8 \text{ m/s}^2$).

This decomposition is the entire foundation. Once you split the initial velocity into horizontal and vertical components, each component obeys simple one-dimensional kinematics — and you already know how to solve those from Uniformly Accelerated Motion.

Setting up the equations

Take a projectile launched from the ground with initial speed u at angle \theta to the horizontal. Place the origin at the launch point, with x pointing horizontally forward and y pointing vertically upward.

Assumptions: The ground is flat. Air resistance is negligible — for a dense object like a cricket ball or a stone over moderate distances, this is a good approximation. The acceleration due to gravity g = 9.8 \text{ m/s}^2 is constant over the height of the trajectory. The Earth's curvature is negligible.

The initial velocity components are:

u_x = u\cos\theta, \qquad u_y = u\sin\theta

Why: resolving the velocity vector along the horizontal and vertical axes. The cosine gives the adjacent component (horizontal), the sine gives the opposite component (vertical). This is the same vector decomposition you met in Velocity and Acceleration in Two Dimensions.

Horizontal motion: constant velocity

There is no horizontal force on the projectile (air resistance is neglected). So the horizontal acceleration is zero, and the horizontal velocity stays constant at u\cos\theta throughout the flight. The horizontal position at time t is:

x(t) = u\cos\theta \cdot t \tag{1}

Why: with zero acceleration, position = initial velocity \times time. This is just s = vt from uniform motion — no complications.

Vertical motion: uniformly accelerated

Gravity acts downward at g = 9.8 \text{ m/s}^2. The vertical component of velocity starts at u\sin\theta (upward) and decreases at rate g. The vertical position at time t is:

y(t) = u\sin\theta \cdot t - \tfrac{1}{2}g\,t^2 \tag{2}

Why: this is the standard kinematic equation s = ut + \frac{1}{2}at^2 applied to the vertical direction. The initial vertical velocity is u\sin\theta, and the acceleration is -g (negative because gravity acts downward, opposite to the positive y-direction). So the \frac{1}{2}at^2 term becomes -\frac{1}{2}g\,t^2.

These two equations — (1) and (2) — completely describe the projectile's position at every instant. Everything else is a consequence.

Watch the projectile fly

The simulation below launches a projectile at 30 \text{ m/s} at 45°. The red dot traces the parabolic path in real time. Notice how the horizontal spacing between trail dots is uniform (constant horizontal speed), while the vertical spacing stretches out near the top (the ball slows down as it rises, then speeds up as it falls).

A projectile launched at $30 \text{ m/s}$ at $45°$ from the ground. The trail shows the parabolic trajectory. Ghost markers appear at $t = 0, 1, 2.165$ (peak), $3$, and $4.33$ s (landing). Click replay to watch again.

The trajectory equation — eliminating time

Equations (1) and (2) give x and y as functions of time t. But often you want the shape of the path — y as a function of x — without worrying about when the projectile is where. You can eliminate t by solving equation (1) for t and substituting into equation (2).

Step 1. From equation (1):

t = \frac{x}{u\cos\theta} \tag{3}

Why: rearrange x = u\cos\theta \cdot t to isolate t. Since u\cos\theta is a constant (horizontal speed never changes), this is just dividing both sides by that constant.

Step 2. Substitute into equation (2):

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

Why: replace every t in equation (2) with the expression from equation (3). This removes time from the picture entirely.

Step 3. Simplify the first term:

u\sin\theta \cdot \frac{x}{u\cos\theta} = x\,\frac{\sin\theta}{\cos\theta} = x\tan\theta

Why: the u cancels, and \sin\theta / \cos\theta = \tan\theta by definition.

Step 4. Simplify the second term:

\frac{1}{2}g \cdot \frac{x^2}{u^2\cos^2\theta} = \frac{gx^2}{2u^2\cos^2\theta}

Why: squaring the fraction x/(u\cos\theta) gives x^2/(u^2\cos^2\theta). Multiply by g/2.

Step 5. Combine:

\boxed{y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}} \tag{4}

Why: this is the equation of the trajectory. It has the form y = ax - bx^2 (where a and b are constants), which is the equation of a parabola opening downward. So a projectile's path is always a parabola — as long as air resistance is negligible and g is constant.

This equation tells you the height of the projectile at any horizontal distance x from the launch point, without needing to know the time. It is the equation a cricketer's sixer traces, the equation a Diwali rocket follows, the equation a water jet from a hose draws in the air.

Time of flight

The projectile starts at ground level (y = 0) and returns to ground level (y = 0). The time of flight T is the total time from launch to landing.

Step 1. Set y(t) = 0 in equation (2):

u\sin\theta \cdot t - \frac{1}{2}g\,t^2 = 0

Why: the projectile is at ground level when y = 0. This happens twice — once at launch and once at landing.

Step 2. Factor out t:

t\left(u\sin\theta - \frac{1}{2}g\,t\right) = 0

Why: both terms contain t. Factoring gives two solutions: t = 0 (the launch, which you already knew) and the other factor set to zero (the landing, which is what you want).

Step 3. Solve the second factor:

u\sin\theta - \frac{1}{2}g\,t = 0

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

\boxed{T = \frac{2u\sin\theta}{g}} \tag{5}

Why: rearrange to isolate t. This is the time of flight — the total time the projectile spends in the air. Notice it depends on u\sin\theta, the vertical component of the initial velocity. A steeper launch angle means a larger vertical component, which means more time in the air.

The time to reach the highest point is exactly half the total flight time: T/2 = u\sin\theta/g. The trajectory is symmetric — the projectile takes the same time to go up as it takes to come down.

Maximum height

At the highest point, the vertical velocity is momentarily zero — the projectile has been decelerating upward, and this is the instant where it stops rising and starts falling.

Step 1. The vertical velocity at time t is:

v_y(t) = u\sin\theta - g\,t

Why: this is v = u + at applied vertically. The initial vertical velocity is u\sin\theta, and the acceleration is -g.

Step 2. At the highest point, v_y = 0:

u\sin\theta - g\,t_{\text{peak}} = 0 \quad \Rightarrow \quad t_{\text{peak}} = \frac{u\sin\theta}{g}

Why: the vertical velocity passes through zero at the peak. Solving gives t_{\text{peak}} = T/2, confirming the symmetry of the trajectory.

Step 3. Substitute t_{\text{peak}} into equation (2):

H = u\sin\theta \cdot \frac{u\sin\theta}{g} - \frac{1}{2}g\left(\frac{u\sin\theta}{g}\right)^2

H = \frac{u^2\sin^2\theta}{g} - \frac{1}{2} \cdot \frac{u^2\sin^2\theta}{g}

\boxed{H = \frac{u^2\sin^2\theta}{2g}} \tag{6}

Why: the first term gives u^2\sin^2\theta/g. The second term gives half of that (because the g in the numerator cancels one power of g from the denominator, leaving u^2\sin^2\theta/(2g)). Subtracting: 1 - 1/2 = 1/2. The maximum height depends on \sin^2\theta — a steeper angle means a greater maximum height, reaching u^2/(2g) for a vertical throw (\theta = 90°).

You can also derive this using the energy method or the kinematic equation v^2 = u^2 + 2as, which gives the same result: 0 = (u\sin\theta)^2 - 2gH, so H = u^2\sin^2\theta/(2g).

Horizontal range

The horizontal range R is the total horizontal distance covered during the flight. Since horizontal velocity is constant:

R = u\cos\theta \cdot T

Why: range = horizontal speed \times total time. The horizontal speed u\cos\theta is constant throughout.

Substitute T = 2u\sin\theta/g:

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

Now use the trigonometric identity 2\sin\theta\cos\theta = \sin 2\theta:

\boxed{R = \frac{u^2\sin 2\theta}{g}} \tag{7}

Why: the double-angle identity compresses 2\sin\theta\cos\theta into \sin 2\theta, making the formula cleaner and revealing something important — the range depends on \sin 2\theta, not on \sin\theta or \cos\theta separately. This single function controls everything about how far the projectile goes.

Maximum range: why 45° is special

For a fixed launch speed u, the range R = u^2\sin 2\theta/g is maximum when \sin 2\theta is maximum. The sine function reaches its maximum value of 1 when its argument is 90°:

\sin 2\theta = 1 \quad \Rightarrow \quad 2\theta = 90° \quad \Rightarrow \quad \theta = 45°

\boxed{R_{\max} = \frac{u^2}{g}} \tag{8}

Why: at 45°, the projectile splits its velocity equally between horizontal and vertical — the perfect balance between going far horizontally and staying in the air long enough vertically. Any steeper angle sends it higher but not as far; any shallower angle keeps it low but it lands too quickly.

For a cricket ball hit at 30 \text{ m/s}, the maximum range (neglecting air resistance) is 30^2 / 9.8 \approx 91.8 \text{ m} — roughly the distance from the pitch to the boundary rope at many Indian cricket grounds.

Complementary angles give equal range

Look at the range formula again: R = u^2\sin 2\theta/g. Consider two angles, \theta and 90° - \theta. The range at angle 90° - \theta is:

R' = \frac{u^2\sin(2(90° - \theta))}{g} = \frac{u^2\sin(180° - 2\theta)}{g}

Since \sin(180° - \alpha) = \sin\alpha:

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

Why: complementary angles \theta and 90° - \theta produce the same value of \sin 2\theta. A projectile launched at 30° travels the same horizontal distance as one launched at 60° with the same speed. The 30° trajectory is low and fast; the 60° trajectory is high and slow. They cover the same range but take different amounts of time and reach different heights.

Explore the angle yourself

The interactive figure below lets you drag the launch angle and watch how the range changes. The launch speed is fixed at 30 \text{ m/s}. Notice the peak at 45° and the symmetry around it — 30° and 60° give the same range, 20° and 70° give the same range.

Drag the red point to change the launch angle. The range (red curve) peaks at $45°$. The dark dot marks the complementary angle ($90° - \theta$) — notice it always sits at the same height as the red dot, confirming that complementary angles give equal range. Launch speed: $30 \text{ m/s}$.

Summary of all the formulas

Here is every result in one place. All assume launch from ground level, with initial speed u at angle \theta, no air resistance, and g = 9.8 \text{ m/s}^2.

Quantity	Formula	What it depends on
Horizontal position	x(t) = u\cos\theta \cdot t	Horizontal speed and time
Vertical position	y(t) = u\sin\theta \cdot t - \tfrac{1}{2}g\,t^2	Vertical speed, gravity, time
Trajectory (shape)	y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}	Parabola in x and y
Time of flight	T = \frac{2u\sin\theta}{g}	Vertical component only
Maximum height	H = \frac{u^2\sin^2\theta}{2g}	Vertical component only
Horizontal range	R = \frac{u^2\sin 2\theta}{g}	Both components (via \sin 2\theta)
Maximum range	R_{\max} = \frac{u^2}{g} at \theta = 45°	Speed only

Notice a pattern: the time of flight and maximum height depend only on the vertical component (u\sin\theta), confirming that the vertical motion is independent. The range involves both components because the projectile needs horizontal speed and time in the air to cover ground.

Worked examples

Example 1: Cricket sixer at 45°

A batsman hits a cricket ball at 30 \text{ m/s} at 45° to the horizontal. Neglecting air resistance, find the time of flight, maximum height, and horizontal range.

Step 1. Identify the knowns.

u = 30 \text{ m/s}, \theta = 45°, g = 9.8 \text{ m/s}^2.

Why: a clean 45° launch is the maximum-range case. The ball is hit hard — 30 \text{ m/s} is about 108 \text{ km/h}.

Step 2. Compute the time of flight.

T = \frac{2u\sin\theta}{g} = \frac{2 \times 30 \times \sin 45°}{9.8} = \frac{2 \times 30 \times 0.7071}{9.8} = \frac{42.43}{9.8}

T \approx 4.33 \text{ s}

Why: at 45°, \sin 45° = \frac{1}{\sqrt{2}} \approx 0.7071. The ball stays in the air for about 4.3 seconds — roughly the time between a batsman hitting a six and the ball landing in the stands.

Step 3. Compute the maximum height.

H = \frac{u^2\sin^2\theta}{2g} = \frac{30^2 \times \sin^2 45°}{2 \times 9.8} = \frac{900 \times 0.5}{19.6} = \frac{450}{19.6}

H \approx 22.96 \text{ m}

Why: \sin^2 45° = (\frac{1}{\sqrt{2}})^2 = 0.5. The ball reaches about 23 \text{ m} — roughly the height of a seven-storey building.

Step 4. Compute the horizontal range.

R = \frac{u^2\sin 2\theta}{g} = \frac{900 \times \sin 90°}{9.8} = \frac{900 \times 1}{9.8}

R \approx 91.84 \text{ m}

Why: at \theta = 45°, \sin 2\theta = \sin 90° = 1, so the range simplifies to u^2/g. The ball lands about 92 \text{ m} from the batsman — past the boundary rope at most Indian grounds.

Result: T \approx 4.33 \text{ s}, H \approx 22.96 \text{ m}, R \approx 91.84 \text{ m}.

The cricket ball's trajectory: launched at $30 \text{ m/s}$ at $45°$, it peaks at $\approx 23 \text{ m}$ (at $t = 2.17 \text{ s}$) and lands at $\approx 92 \text{ m}$ (at $t = 4.33 \text{ s}$). Click replay to watch again.

What this shows: At 45°, the ball achieves maximum range for this speed. The trajectory is a symmetric parabola — the rising half mirrors the falling half. The time to reach the peak (2.17 \text{ s}) is exactly half the total flight time (4.33 \text{ s}).

Example 2: Complementary angles — 30° vs 60°

An ISRO sounding rocket test-fires two identical rockets at 20 \text{ m/s}, one at 30° and the other at 60°, to verify the complementary angle principle. Find the range, maximum height, and time of flight for each.

Step 1. Compute values for \theta = 30°.

T_{30} = \frac{2 \times 20 \times \sin 30°}{9.8} = \frac{2 \times 20 \times 0.5}{9.8} = \frac{20}{9.8} \approx 2.04 \text{ s}

H_{30} = \frac{20^2 \times \sin^2 30°}{2 \times 9.8} = \frac{400 \times 0.25}{19.6} = \frac{100}{19.6} \approx 5.10 \text{ m}

R_{30} = \frac{20^2 \times \sin 60°}{9.8} = \frac{400 \times 0.8660}{9.8} \approx \frac{346.4}{9.8} \approx 35.35 \text{ m}

Why: at 30°, the vertical component is 20 \times 0.5 = 10 \text{ m/s} and the horizontal component is 20 \times 0.866 = 17.3 \text{ m/s}. Most of the speed goes into horizontal travel, so the ball stays low but moves fast sideways.

Step 2. Compute values for \theta = 60°.

T_{60} = \frac{2 \times 20 \times \sin 60°}{9.8} = \frac{2 \times 20 \times 0.8660}{9.8} = \frac{34.64}{9.8} \approx 3.53 \text{ s}

H_{60} = \frac{20^2 \times \sin^2 60°}{2 \times 9.8} = \frac{400 \times 0.75}{19.6} = \frac{300}{19.6} \approx 15.31 \text{ m}

R_{60} = \frac{20^2 \times \sin 120°}{9.8} = \frac{400 \times 0.8660}{9.8} \approx 35.35 \text{ m}

Why: at 60°, the vertical component is 20 \times 0.866 = 17.3 \text{ m/s} and the horizontal component is 20 \times 0.5 = 10 \text{ m/s}. The roles are reversed — the ball goes high but not as fast horizontally. Yet it stays in the air 1.5 seconds longer, which exactly compensates for the lower horizontal speed.

Step 3. Compare.

	\theta = 30°	\theta = 60°
Time of flight	2.04 \text{ s}	3.53 \text{ s}
Maximum height	5.10 \text{ m}	15.31 \text{ m}
Range	35.35 \text{ m}	35.35 \text{ m}

Why: the ranges are identical — as the complementary angle theorem predicts. The 60° projectile goes three times higher and stays in the air 73\% longer, but covers the same ground distance. The product (horizontal speed \times time) is the same for both.

Result: Both rockets land at 35.35 \text{ m}. The 30° rocket stays low (5.1 \text{ m} peak) and fast; the 60° rocket soars to 15.3 \text{ m} but takes longer.

Two projectiles at $20 \text{ m/s}$: the red one at $30°$ (low, fast) and the dark one at $60°$ (high, slow). Both land at the same distance ($35.35 \text{ m}$). The $30°$ projectile lands first — it takes $2.04 \text{ s}$ versus $3.53 \text{ s}$ for the $60°$ one. Click replay to compare.

What this shows: Complementary angles always give equal range. The 30° trajectory sacrifices height for horizontal speed; the 60° trajectory sacrifices horizontal speed for time in the air. The trade-off is perfectly balanced — the range comes out the same.

Common confusions

"The projectile slows down at the top." Partially true, partially misleading. The vertical velocity is zero at the peak — the projectile momentarily stops rising. But the horizontal velocity is u\cos\theta throughout, including at the peak. The projectile never stops moving. Its speed is minimum at the top (equal to u\cos\theta, the horizontal component), not zero.
"A steeper angle always means a higher range." No. A steeper angle increases the time of flight (more u\sin\theta) but decreases horizontal speed (u\cos\theta). The range is horizontal speed \times time, and the optimal trade-off is at 45°. Beyond 45°, the loss in horizontal speed outweighs the gain in time aloft.
"The formulas work with air resistance." They do not. Air resistance adds a horizontal deceleration (the projectile slows down in the air), makes the trajectory asymmetric (steeper on the descent), and reduces the range significantly. For a shuttlecock, air resistance dominates and the trajectory looks nothing like a parabola. For a dense cricket ball over 60-80 \text{ m}, the parabolic approximation is reasonable but not exact.
"45° always gives maximum range." Only when launched from flat ground. When launched from a height (off a cliff, from a building roof), the optimal angle is less than 45°. When launched up a slope, the optimal angle depends on the slope angle. The 45° result assumes launch and landing are at the same height.
"Complementary angles give equal everything." No — they give equal range only. The time of flight, maximum height, and shape of the trajectory are all different. The steeper angle always gives the greater height and longer flight time.

If you came here to understand projectile motion, use the formulas, and solve problems, you have everything you need. What follows is for readers who want the connection to the trajectory equation's geometry and the JEE-level analysis of how the formulas interrelate.

The trajectory equation as a downward parabola

The trajectory equation (4) can be written as:

y = x\tan\theta - \frac{g}{2u^2\cos^2\theta}\,x^2

Let A = \tan\theta and B = \frac{g}{2u^2\cos^2\theta}. Then y = Ax - Bx^2, which is a parabola opening downward (since B > 0).

The vertex of this parabola is at:

x_{\text{vertex}} = \frac{A}{2B} = \frac{\tan\theta}{2 \cdot \frac{g}{2u^2\cos^2\theta}} = \frac{u^2\sin\theta\cos\theta}{g} = \frac{u^2\sin 2\theta}{2g} = \frac{R}{2}

Why: the vertex of y = Ax - Bx^2 is at x = A/(2B). This gives x = R/2 — the peak is exactly at the midpoint of the range, confirming the symmetry of the parabolic trajectory. The ball reaches its highest point when it has covered exactly half the horizontal distance.

The y-coordinate of the vertex is:

y_{\text{vertex}} = \frac{A^2}{4B} = \frac{\tan^2\theta}{4 \cdot \frac{g}{2u^2\cos^2\theta}} = \frac{u^2\sin^2\theta}{2g} = H

This confirms that the maximum height H from equation (6) is indeed the vertex height.

Range-height relation

From the formulas, you can show that:

R = \frac{4H}{\tan\theta}

Derivation:

\frac{4H}{\tan\theta} = \frac{4 \cdot \frac{u^2\sin^2\theta}{2g}}{\frac{\sin\theta}{\cos\theta}} = \frac{2u^2\sin^2\theta}{g} \cdot \frac{\cos\theta}{\sin\theta} = \frac{2u^2\sin\theta\cos\theta}{g} = \frac{u^2\sin 2\theta}{g} = R

Why: this relation connects range and height through the angle. It means that for a given height, a lower angle gives a longer range. JEE problems sometimes give you H and \theta and ask for R — this formula gives it directly.

The time-of-flight and height connection

Since T = 2u\sin\theta/g and H = u^2\sin^2\theta/(2g), you can eliminate u\sin\theta:

From T: u\sin\theta = gT/2

Substitute into H: H = \frac{(gT/2)^2}{2g} = \frac{g^2T^2}{4 \cdot 2g} = \frac{gT^2}{8}

\boxed{H = \frac{gT^2}{8}}

Why: this result is independent of u and \theta — once you know the flight time, the maximum height is fixed at gT^2/8. Conversely, T = \sqrt{8H/g}. JEE Advanced problems love asking for one quantity when you know the other.

What changes if you add air resistance?

With air resistance, the physics changes qualitatively:

The trajectory is asymmetric. The descending half is steeper than the ascending half because the projectile is slower on the way down (drag reduced the speed throughout the flight).
The range decreases. Drag removes kinetic energy continuously, so the projectile does not travel as far.
The optimal angle is less than 45°. With drag, it pays to launch at a lower angle (roughly 30°-40° for typical conditions) because a lower angle means less time in the air and therefore less speed lost to drag.
The maximum height is reduced. Drag opposes the upward motion, so the projectile does not climb as high.

The full analysis requires solving a differential equation with a velocity-dependent drag force: \vec{F}_{\text{drag}} = -b\vec{v} (linear drag) or \vec{F}_{\text{drag}} = -c|\vec{v}|\vec{v} (quadratic drag). The quadratic model is more realistic for objects like cricket balls at normal speeds, but it has no closed-form solution — you need numerical methods. This is covered in Projectile Motion — Advanced Problems.

Where this leads next

Projectile Motion — Advanced Problems — projectiles on inclined planes, launched from a height, relative motion between projectiles, and the effect of air resistance.
Free Fall — the special case \theta = 90°: vertical motion under gravity alone.
Relative Motion — two projectiles launched simultaneously appear to move in a straight line relative to each other. This is a powerful problem-solving technique.
Uniformly Accelerated Motion — the kinematic equations that underpin the vertical component of projectile motion.
Velocity and Acceleration in Two Dimensions — the vector decomposition that makes the two-component treatment possible.