05Motion in Two Dimensions
Extend kinematics to two dimensions: resolve vectors into components, analyze projectile motion, and solve problems involving objects launched at angles.
Real-world motion rarely stays in a straight line. Projectile motion and vector decomposition are essential for understanding everything from thrown balls to satellite orbits.
How can we describe and predict the motion of objects that move in two dimensions, such as a ball thrown through the air or a car turning a corner?
Lesson Overview
In one-dimensional kinematics we described motion along a straight line. Real-world motion, however, is almost always two-dimensional. A soccer ball kicked at an angle, a car navigating a curve, or a satellite in orbit all require us to track both horizontal and vertical (or x and y) components simultaneously. In this lesson we extend our kinematics toolkit by introducing vectors — quantities that have both magnitude and direction — and learning how to break them into perpendicular components. We then apply these ideas to projectile motion, discovering that the horizontal and vertical components of a projectile's velocity are completely independent of each other. This independence principle is one of the most powerful ideas in classical mechanics.
Vector Components
Any vector A with magnitude A at angle θ above the positive x-axis can be resolved into two perpendicular components:
Horizontal component
Aₓ = A cos θ
Vertical component
Aᵧ = A sin θ
Resultant magnitude
R = √(Rₓ² + Rᵧ²)
Resultant direction
θ = arctan(Rᵧ / Rₓ)
Tip-to-tail method: Draw vectors head-to-tail; the resultant runs from the tail of the first to the head of the last. The component method is more precise: add all x-components to get Rₓ, add all y-components to get Rᵧ, then find R and θ.
Projectile Motion
Horizontal motion
Constant velocity — no horizontal force acts on the projectile (ignoring air resistance).
x = v₀ₓ t
Vertical motion
Free fall — gravity accelerates the object downward at g = 9.8 m/s².
y = v₀ᵧ t − ½g t²
vᵧ = v₀ᵧ − g t
Independence principle
The horizontal and vertical motions of a projectile are completely independent. Horizontal velocity never changes; vertical velocity changes at rate g. This means a ball dropped straight down and a ball thrown horizontally from the same height hit the ground at the same time.
Key Equations
x-component
Aₓ = A cos θ
y-component
Aᵧ = A sin θ
Resultant magnitude
R = √(Rₓ² + Rᵧ²)
Horizontal position
x = v₀ₓ t
Vertical position
y = v₀ᵧ t − ½g t²
Vertical velocity
vᵧ = v₀ᵧ − g t
Range equation
R = v₀² sin 2θ / g
Worked Examples
A force vector has a magnitude of 50 N and points at 37° above the positive x-axis. Find its x- and y-components.
Identify: A = 50 N, θ = 37°
Aₓ = A cos θ = 50 × cos 37° = 50 × 0.7986 ≈ 39.9 N ≈ 40 N
Aᵧ = A sin θ = 50 × sin 37° = 50 × 0.6018 ≈ 30.1 N ≈ 30 N
Check: √(40² + 30²) = √(1600 + 900) = √2500 = 50 N ✓
Vector A = 30 m due East and vector B = 40 m due North. Find the magnitude and direction of the resultant vector R = A + B.
Set up coordinates: East = +x, North = +y
Rₓ = Aₓ + Bₓ = 30 + 0 = 30 m
Rᵧ = Aᵧ + Bᵧ = 0 + 40 = 40 m
R = √(Rₓ² + Rᵧ²) = √(30² + 40²) = √(900 + 1600) = √2500 = 50 m
θ = arctan(Rᵧ / Rₓ) = arctan(40 / 30) = arctan(1.333) ≈ 53.1°
A ball is launched horizontally at 20 m/s from the top of a 45 m cliff. Find the time of flight and the horizontal range.
Horizontal: v₀ₓ = 20 m/s, v₀ᵧ = 0 (launched horizontally)
Vertical: y = v₀ᵧ t − ½g t² → −45 = 0 − ½(9.8)t²
45 = 4.9 t² → t² = 45/4.9 = 9.184 → t = √9.184 ≈ 3.03 s
Horizontal range: x = v₀ₓ t = 20 × 3.03 ≈ 60.6 m
A projectile is launched at 30 m/s at 45° above the horizontal. Find the maximum height and the total horizontal range.
Resolve initial velocity: v₀ₓ = 30 cos 45° ≈ 21.2 m/s; v₀ᵧ = 30 sin 45° ≈ 21.2 m/s
Max height: at peak vᵧ = 0 → 0 = v₀ᵧ − g t_peak → t_peak = 21.2/9.8 ≈ 2.16 s
H = v₀ᵧ t_peak − ½g t_peak² = 21.2(2.16) − ½(9.8)(2.16²) ≈ 45.8 − 22.9 ≈ 22.9 m
Range: R = v₀² sin 2θ / g = 30² × sin 90° / 9.8 = 900/9.8 ≈ 91.8 m
A ball is kicked at 25 m/s at 30° above the horizontal. Find the initial velocity components and the time to reach maximum height.
v₀ₓ = 25 cos 30° = 25 × 0.866 ≈ 21.7 m/s
v₀ᵧ = 25 sin 30° = 25 × 0.500 = 12.5 m/s
At max height vᵧ = 0: 0 = v₀ᵧ − g t → t = v₀ᵧ / g = 12.5 / 9.8 ≈ 1.28 s
Guided Problems
A displacement vector has magnitude 65 m at 53° above the positive x-axis. Find its x- and y-components.
Hint: Use Aₓ = A cos θ and Aᵧ = A sin θ. cos 53° ≈ 0.602, sin 53° ≈ 0.799.
Two vectors: P = 12 m East and Q = 16 m South. Find the magnitude and direction of P + Q.
Hint: Let East = +x, North = +y. Q points South so Qᵧ = −16 m. Use R = √(Rₓ² + Rᵧ²) and θ = arctan(|Rᵧ|/|Rₓ|) below the x-axis.
A stone is thrown horizontally at 15 m/s from a bridge 20 m above a river. How far from the base of the bridge does it land?
Hint: First find the time to fall 20 m using y = ½g t² (v₀ᵧ = 0). Then use x = v₀ₓ t.
A soccer ball is kicked at 18 m/s at 40° above the horizontal. What is the ball's speed at the highest point of its trajectory?
Hint: At the highest point, the vertical velocity is zero. Only the horizontal component remains: v = v₀ₓ = v₀ cos 40°.
A cannonball is fired at 60 m/s at 60° above the horizontal. Use the range equation to find how far it travels horizontally before landing.
Hint: R = v₀² sin 2θ / g. Note that sin(2 × 60°) = sin 120° = sin 60° ≈ 0.866.
Key Vocabulary
Vector
A quantity that has both magnitude (size) and direction, represented by an arrow.
Example: Velocity 30 m/s North; Force 50 N at 37°
Scalar
A quantity that has magnitude only, with no direction associated.
Example: Speed 30 m/s; Temperature 25°C; Mass 5 kg
Component
The projection of a vector onto a coordinate axis. Every 2-D vector has an x-component and a y-component.
Example: For A = 50 N at 37°: Aₓ = 40 N, Aᵧ = 30 N
Resultant
The single vector that represents the combined effect of two or more vectors added together.
Example: 30 m East + 40 m North → resultant 50 m at 53.1° N of E
Projectile
Any object that is launched into the air and moves under the influence of gravity alone (no thrust or air resistance).
Example: A thrown ball, a kicked soccer ball, a bullet
Trajectory
The curved path followed by a projectile through space; a parabola when gravity is the only force.
Example: The arc traced by a basketball from release to the hoop
Range
The total horizontal distance traveled by a projectile from launch to landing on the same horizontal level.
Example: R = v₀² sin 2θ / g; maximum range occurs at θ = 45°
Independence of motion
The principle that horizontal and vertical motions of a projectile are completely independent; each is governed by its own equations.
Example: A dropped ball and a horizontally thrown ball from the same height land simultaneously.
Workbook Check — Interactive Quiz
Interactive Practice — 5 Questions
A vector has magnitude 100 m at 30° above the x-axis. What is its x-component?
A ball is launched horizontally from a height of 80 m. Approximately how long does it take to reach the ground? (g = 9.8 m/s²)
Which launch angle gives the maximum horizontal range for a projectile on level ground?
At the highest point of its trajectory, a projectile launched at an angle has:
Two vectors: A = 6 m East and B = 8 m North. What is the magnitude of A + B?
Independent Practice
A ball is kicked horizontally at 15 m/s from a cliff 20 m high. Find (a) the time of flight and (b) the horizontal range. (g = 9.8 m/s²)
A projectile is launched at 30° above horizontal with an initial speed of 40 m/s. Find the maximum height reached. (g = 9.8 m/s²)
A boat can travel at 4.0 m/s in still water. It heads directly across a 120 m wide river with a current of 3.0 m/s. Find (a) the resultant speed and (b) how far downstream it lands.
A car rounds a flat circular curve of radius 80 m at 20 m/s. Calculate the centripetal acceleration and the minimum coefficient of static friction needed. (g = 9.8 m/s²)
★ A soccer ball is kicked at 45° above horizontal and lands 60 m away on level ground. (a) Find the initial speed. (b) Find the maximum height. (c) Find the speed of the ball at the highest point. (d) Find the velocity (magnitude and direction) just before it lands. (g = 9.8 m/s²)
ChallengeCommon Mistakes
Thinking horizontal velocity changes during projectile flight
Horizontal velocity is constant (no air resistance); only vertical velocity changes due to gravity
Using the full launch speed v₀ instead of its components in kinematic equations
Always resolve v₀ into v₀ₓ = v₀cosθ and v₀ᵧ = v₀sinθ before applying any kinematic equation
Forgetting that time is shared between horizontal and vertical motion
The same time t applies to both x and y directions. Find t from the vertical equation, then use it in the horizontal equation
Adding relative velocities as if direction doesn't matter
Relative velocity is a vector sum: v_AC = v_AB + v_BC. Draw a vector diagram and use components or the law of cosines
Math Tips
Projectile range formula: R = v₀²sin(2θ)/g. Maximum range occurs at θ = 45°. Complementary angles (e.g., 30° and 60°) give the same range
Time of flight for a projectile launched and landing at the same height: T = 2v₀sinθ/g. The object reaches max height at T/2
To find the resultant of two perpendicular vectors A and B: magnitude = √(A² + B²), angle = arctan(B/A). Always draw the vector triangle first
Uniform circular motion: centripetal acceleration a_c = v²/r = ω²r always points toward the center. Speed is constant but velocity direction changes continuously