AP Physics C: Mechanics · Unit 1: Kinematics · Lesson 1.5

Deep Dive: Motion in Two or Three Dimensions

🔬 Deep Dive

This is your textbook for this topic. Take your time. Read it more than once.

1.5.A.11.5.A.21.5.A.3Concept

Splitting Motion Into Independent Components

Motion in two or three dimensions can be analyzed using the one-dimensional kinematic relationships you already know — as long as you separate the motion into components first. Velocity and acceleration can be completely different, and change in completely different ways, along each axis. Most importantly: changing an object's motion in one dimension has zero effect on a perpendicular dimension.

🔑This independence is the single idea that makes 2D motion tractable. Instead of one hard 2D problem, you get two easy 1D problems that happen to share a clock. Solve each one separately using everything from Lessons 1.2 and 1.3, then recombine at the end if you need a magnitude or direction.

1.5.A.4ConceptMath

Projectile Motion — The Special Case

Projectile motion is 2D motion with a very specific pattern: zero acceleration in one direction and constant, nonzero acceleration in the other. Near Earth's surface, that means horizontal acceleration is zero and vertical acceleration is −g.

Horizontal motion — constant velocity

vₓ(t) = vₓ₀   (constant)

x(t) = x₀ + vₓ₀t

Vertical motion — constant acceleration

vy(t) = vy₀ − gt

y(t) = y₀ + vy₀t − ½gt²

⚠️The single most common projectile-motion mistake: letting the horizontal and vertical equations "leak" into each other. The horizontal velocity never appears in a vertical equation, and vice versa — the only thing the two directions share is the same value of t.

If the projectile is launched at an angle θ with initial speed v₀, decompose it first using exactly the unit-vector-notation skills from Lesson 1.1:

vₓ₀ = v₀ cos θ

vy₀ = v₀ sin θ

Math

Solving Projectile Motion Problems

Three quantities come up in almost every projectile problem: how long it's in the air, how high it gets, and how far it travels. All three fall directly out of the vertical and horizontal equations above.

Maximum height

The projectile is at its peak when its vertical velocity is momentarily zero — exactly like the flat tangent line you saw in Lesson 1.3.

t(peak) = vy₀ / g

y(max) = y₀ + vy₀² / (2g)

Time of flight

Set y(t) = 0 (or whatever the landing height is) and solve the resulting quadratic for the positive root.

Range

Once you know the time of flight, plug it into the horizontal position equation — horizontal velocity never changed, so this step is just multiplication.

Range = vₓ₀ × t(flight)

Explore all of this directly below: launch a projectile with your own speed, angle, and starting height, then scrub through its flight to see the horizontal and vertical velocity components behaving completely independently, exactly as 1.5.A.1–.3 describe.

Launch a projectile. Set its initial speed, angle, and starting height, then scrub the time slider to track its position along the flight path — watch how the horizontal velocity (blue) never changes while the vertical velocity (violet) constantly does.

Launch speed v₀18 m/s

Launch angle θ50°

Launch height y₀0 m

time t0.00 s

vx = 11.6 m/s (constant)vy = 13.8 m/s

Time of flight = 2.76 sMax height = 9.5 m (at t = 1.38 s)Range = 31.9 m

ExampleGuided Example — Projectile Launched from the Ground

A ball is launched from ground level at 20 m/s, at an angle of 40° above the horizontal. Find its time of flight, maximum height, and range. Use g = 10 m/s².

Step 1 — Decompose the initial velocity

vₓ₀ = 20 cos 40° ≈ 15.3 m/s · vy₀ = 20 sin 40° ≈ 12.9 m/s

ExampleWorked Example — Projectile Launched from a Height

A rock is thrown horizontally at 12 m/s from the top of a 45 m cliff. Find its time of flight and how far from the base of the cliff it lands.

← Back to Lesson 1.5That wraps up Unit 1! Unit 2 moves from describing motion to explaining it — forces and Newton's laws.