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Vectors and Motion
in Two Dimensions

Master the art of breaking motion into components, adding vectors, and solving projectile motion problems with confidence.

AP Physics 1

What Is a Vector?

A vector is a quantity that has both magnitude (size) and direction. Unlike a scalar (which only has magnitude), a vector tells you not just "how much" but also "which way."

Vector Examples
Displacement, velocity, acceleration, force
Scalar Examples
Speed, mass, time, temperature, energy

Representing Vectors Graphically

x y \(\vec{A}\) \(A_x\) \(A_y\) \(\theta\)

Vector Component Equations

Given a vector \(\vec{A}\) with magnitude \(A\) at angle \(\theta\) from the positive x-axis:

x-Component
$$A_x = A\cos\theta$$
y-Component
$$A_y = A\sin\theta$$

Adding Vectors: Tip-to-Tail Method

\(\vec{A}\) \(\vec{B}\) \(\vec{R} = \vec{A}+\vec{B}\)

Place the tail of \(\vec{B}\) at the tip of \(\vec{A}\). The resultant \(\vec{R}\) goes from the tail of \(\vec{A}\) to the tip of \(\vec{B}\).

Projectile Motion: The Big Idea

A projectile is any object moving through the air under only the influence of gravity. The key insight: horizontal and vertical motions are independent.

Horizontal (x)
Constant velocity: \(a_x = 0\)
Vertical (y)
Constant acceleration: \(a_y = -g\)

The Core Kinematic Equations

Horizontal
$$x = x_0 + v_{0x}\,t$$ $$v_x = v_{0x} = v_0\cos\theta$$
Vertical
$$y = y_0 + v_{0y}\,t - \tfrac{1}{2}g\,t^2$$ $$v_y = v_{0y} - g\,t$$

Trajectory of a Projectile

\(\vec{v}_0\) \(v_x\) only \(\vec{v}_f\) Maximum Height \(h_{max}\) Range \(R\) \(\theta\)

Scalars vs Vectors: Key Distinction

Scalar Vector
Has magnitude? ✓ Yes ✓ Yes
Has direction? ✕ No ✓ Yes
Addition rule Arithmetic Tip-to-tail / components
Example Speed: \(5\;\text{m/s}\) Velocity: \(5\;\text{m/s}\) east
Can be negative? Only some (temp) Components can be

How to Solve 2D Motion Problems

1
Draw a diagram. Label known values, choose a coordinate system, and identify the angle \(\theta\).
2
Resolve into components. Find \(v_{0x} = v_0\cos\theta\) and \(v_{0y} = v_0\sin\theta\).
3
Write x- and y-equations separately. Use horizontal (constant velocity) and vertical (constant acceleration) kinematics.
4
Link with time. Find \(t\) from one dimension, substitute into the other to solve for the unknown.
5
Check units and signs. Verify the answer makes physical sense (e.g., positive range, reasonable time).

Example 1: Finding Components

A ball is launched at \(v_0 = 20\;\text{m/s}\) at \(\theta = 30^\circ\) above the horizontal. Find the initial horizontal and vertical velocity components.

\(20\;\text{m/s}\) \(v_{0x}\) \(v_{0y}\) \(30^\circ\)

Example 2: Time of Flight

Using the same ball (\(v_0 = 20\;\text{m/s}\), \(\theta = 30^\circ\)), find the total time of flight if it lands at the same height it was launched.

1
At landing, \(y = 0\). Use: \(0 = v_{0y}\,t - \frac{1}{2}g\,t^2\)
2
Factor: \(t\!\left(v_{0y} - \frac{1}{2}g\,t\right) = 0\)
3
Non-zero solution: \(t = \dfrac{2\,v_{0y}}{g} = \dfrac{2 \times 10.0}{9.8}\)

Example 3: Range and Max Height

Still using \(v_0 = 20\;\text{m/s}\), \(\theta = 30^\circ\), \(t = 2.04\;\text{s}\). Find the range and maximum height.

Range
\(R = v_{0x} \times t\)
\(R = 17.3 \times 2.04\)
\(R = 35.3\;\text{m}\)
Max Height
At peak: \(v_y = 0\), so \(t_{peak} = \frac{v_{0y}}{g} = 1.02\;\text{s}\)
\(h = v_{0y}\,t_{peak} - \frac{1}{2}g\,t_{peak}^2\)
\(h = 5.10\;\text{m}\)

Special Case: Horizontal Launch

\(v_{0x}\) \(h\) Range \(R\)

A relief package meeting a target below a moving helicopter

A helicopter and released package above a canyon accompany an exact no-drag trajectory with equal-time positions and impact distance.The package inherits the helicopter’s horizontal velocity at release. Gravity changes its vertical velocity while both components evolve over the same elapsed time.

Horizontal release from a helicopter

Horizontal release from a helicopter. For an 80 m release height, horizontal speed 20 m/s and g = 9.8 m/s², flight time is about 4.04 s and range is about 80.81 m. Without drag, the package and helicopter retain the same horizontal velocity.For an 80 m release height, horizontal speed 20 m/s and g = 9.8 m/s², flight time is about 4.04 s and range is about 80.81 m. Without drag, the package and helicopter retain the same horizontal velocity.

Common Mistakes to Avoid

✕ Mixing Components
Using \(v_0\) instead of \(v_{0x}\) or \(v_{0y}\) in a single-dimension equation. Always decompose first.
✕ Wrong Trig Function
Using \(\sin\) when you need \(\cos\) (or vice versa). Remember: \(\cos\) is adjacent (horizontal), \(\sin\) is opposite (vertical).
✕ Forgetting Signs
Gravity is negative (\(a_y = -9.8\;\text{m/s}^2\)) if you choose up as positive. Inconsistent signs ruin results.

Summary: Key Formulas at a Glance

Quantity Horizontal (x) Vertical (y)
Acceleration \(a_x = 0\) \(a_y = -g\)
Initial velocity \(v_0\cos\theta\) \(v_0\sin\theta\)
Velocity at time \(t\) \(v_{0x}\) (constant) \(v_{0y} - g\,t\)
Position at time \(t\) \(x_0 + v_{0x}\,t\) \(y_0 + v_{0y}\,t - \frac{1}{2}g\,t^2\)
Time of flight \(t = \dfrac{2\,v_0\sin\theta}{g}\) (same-height)
Range \(R = \dfrac{v_0^2\sin 2\theta}{g}\) (same-height)

Question 1 of 5

A vector has magnitude \(10\;\text{m/s}\) and points at \(60^\circ\) above the horizontal. What is its horizontal component?

Question 2 of 5

A projectile is launched at \(45^\circ\) with a speed of \(20\;\text{m/s}\). At the highest point of its trajectory, what is the vertical component of velocity?

Question 3 of 5

Two balls are released simultaneously from the same height: one is dropped straight down, the other is launched horizontally. Which hits the ground first?

Question 4 of 5

A ball is thrown horizontally at \(15\;\text{m/s}\) from a cliff that is \(80\;\text{m}\) tall. How far from the base of the cliff does it land? (Use \(g = 10\;\text{m/s}^2\))

Question 5 of 5

For a projectile launched from flat ground, at which angle is the range maximized?