AP Physics 1  ·  Unit 1: Kinematics  ·  Lesson 1.5

Deep Dive: Vectors and Motion in Two Dimensions

🔬 Deep Dive
This is your textbook for this topic. Take your time. Read it more than once.
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Any Vector Is Two Perpendicular Vectors

Here's the core idea of this entire lesson: any vector — no matter what angle it points — can be thought of as the result of adding two simpler vectors that are perpendicular to each other. One points purely horizontal, one points purely vertical. Together, they add up to the original vector.

🔑This isn't a trick or an approximation. It's mathematically exact. A vector pointing 37° above the ground is identical to a horizontal vector plus a vertical vector of the right sizes, added together.
xyv = 50 m/sθ = 37°vₓ = v cos θ = 40 m/svy = v sin θ = 30 m/s

The horizontal piece is called the x-component. The vertical piece is the y-component. Resolving a vector means finding those two component values.

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Choosing a Coordinate System

Before resolving any vector, you choose a coordinate system — which direction is positive x, which is positive y. This choice is yours to make, and a smart choice can turn a hard problem into an easy one.

💡For projectile motion, the standard choice is horizontal = x-axis, vertical = y-axis, with up as positive. That choice isn't required by physics — it's just convenient because gravity acts purely along one axis that way.

Once you've picked your axes, every vector in the problem gets resolved relative to that coordinate system — consistently, for the whole problem.

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Trigonometry Does the Work

If you know a vector's magnitude and the angle it makes with the x-axis, three trig relationships let you find both components:

vₓ = v cos θ
vy = v sin θ
v = √(vₓ² + vy²)

The first two break a vector into components. The third — the Pythagorean theorem — does the reverse: combines components back into the original magnitude. They're inverses of each other.

Adjust the magnitude and angle. Watch the vector resolve into its x and y components live.

Magnitude (v)50 m/s
Angle (θ)37°
vₓ = 39.9vy = 30.1
vₓ = v cos θ = 39.93 m/svy = v sin θ = 30.09 m/scheck: √(vₓ²+vy²) = 50.00 m/s
ExampleWorked Example — Resolving a Velocity Vector

A ball is kicked at 25 m/s at an angle of 30° above the ground. Find the horizontal and vertical components of its velocity.

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Splitting 2D Motion into Two 1D Problems

Once a vector is resolved into components, something powerful happens: the horizontal motion and the vertical motion become completely independent. Whatever happens in the x-direction has zero effect on what happens in the y-direction, and vice versa.

🔑This means every kinematic equation you learned in Lesson 1.3 still works — you just apply it twice. Once for x. Once for y. Using the same equations, just with different values plugged in for each axis.

This is why 2D motion in AP Physics 1 never actually requires new equations. You already know everything you need. The only new skill is realizing you have to do the same work twice — once per axis — and keep the two completely separate until the very end.

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Projectile Motion — The Classic Case

A projectile is anything launched into the air and left to move under gravity alone — a thrown ball, a launched rocket (before its engine cuts off), a kicked soccer ball. Projectile motion is the most common 2D motion problem in AP Physics 1, and it has a defining feature:

Horizontal (x)

Zero acceleration. Velocity stays constant the entire flight. No force acts horizontally (ignoring air resistance).

Vertical (y)

Constant acceleration: g ≈ 10 m/s² downward. Velocity changes continuously — slows going up, speeds up coming down.

⚠️The classic mistake: using the same time, velocity, or acceleration value for both axes. They are governed by completely different rules. The only thing horizontal and vertical motion share is the same total time in the air — because it's the same object, landing at the same moment.

Launch a projectile. Drag the time slider and watch horizontal velocity stay constant while vertical velocity changes — two independent 1D problems, happening at once.

Launch speed40 m/s
Launch angle45°
Time (t)0.00 s
vₓ = 28.3 m/s (constant)vy = 28.3 m/s (changing)height = 0.0 m

Notice: vₓ never changes as you drag the time slider. vy starts positive, hits zero at the peak, then goes negative.

ExampleGuided Example — Projectile Launched at an Angle

A ball is launched at 30 m/s at 53° above the horizontal. Find the time it spends in the air and the horizontal distance it travels (use g = 10 m/s²).

Step 1Resolve the initial velocity
vₓ = v cos(53°) = 30 × 0.602 = 18.06 m/s
vy = v sin(53°) = 30 × 0.799 = 23.97 m/s
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