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AP Physics 1: Projectile Motion & Vectors

The “Big Idea”: Projectile motion is just two simple 1D motion problems happening at the same time. The horizontal and vertical motions represent independent realities.

1. Breaking Down Vectors

Before solving projectile problems, you must be able to split a diagonal velocity vector into its components. In AP Physics 1, we use trigonometry (SOH CAH TOA) to do this.

A right-angled triangle diagram showing a velocity vector 'v' at angle theta being resolved into horizontal component v_x and vertical component v_y. — Resolving an initial velocity vector ( $v$ ) into its horizontal ( $v_x$ ) and vertical ( $v_y$ ) components using trigonometry.

Horizontal Component (v_x):
$v_x = v \cos(\theta)$
Vertical Component (v_y):
$v_y = v \sin(\theta)$

Calculus Note: AP Physics 1 does not use unit vectors ( $\hat{i}, \hat{j}, \hat{k}$ ) or dot products. Stick to x and y components.

2. The Two Rules of Projectile Motion

If you memorize nothing else, memorize this table. This is how you set up every FRQ.

Axis	Acceleration	Velocity Behavior	Equation to Use
Horizontal (X)	$a_x = 0$	Constant Velocity	$\Delta x = v_x t$
Vertical (Y)	$a_y = -9.8 \, \text{m/s}^2$	Changing (Free Fall)	Use “Big 3” Kinematics Equations

The Bridge: The only variable that is the same for both X and Y sides is Time (t).

A diagram of a projectile's parabolic path showing that horizontal velocity vectors remain constant length while vertical velocity vectors change due to gravity. — The independence of motion: Notice how the horizontal blue arrows stay the same length, while the vertical red arrows change due to gravitational acceleration ( $g$ ).

3. Three Scenarios You Will See

A. Horizontal Launch

Object thrown straight off a cliff.

Initial $v_{0y} = 0$
Initial $v_{0x} = v_{launch}$
Time depends only on height!

B. Angled Launch (Ground-to-Ground)

Object kicked like a soccer ball.

At the peak height, $v_y = 0$ (but $v_x$ is still there!)
Time up = Time down (if landing at same height).

4. AP-Style Concept Check

Try this “Paragraph Length Response” style question. No numbers allowed!

Question: Two identical balls are released from the top of a cliff at the same time. Ball A is dropped from rest. Ball B is thrown horizontally outward with speed $v$ . Which ball hits the ground first? Justify your answer.

Click to see the Answer

Answer: They hit at the same time.

Reasoning: The vertical motion of an object is independent of its horizontal motion. Both balls start with an initial vertical velocity of zero ( $v_{0y} = 0$ ) and fall the same vertical distance ( $\Delta y$ ) under the same acceleration due to gravity ( $g$ ). Therefore, according to the equation $\Delta y = v_{0y}t + \frac{1}{2}gt^2$ , the time $t$ to fall must be identical for both.

5. AP-Style Derivation Practice

On the AP Exam, you are often asked to derive equations using only variables. Practice these three common scenarios. Do not memorize the final answers; memorize the steps!

Derivation 1: Total Horizontal Range ( $R$ )
Find the horizontal distance derived in terms of $v_0$ , $\theta$ , and $g$ .

Click to see Step-by-Step Derivation

Step 1: Horizontal Motion
$R = v_{0x} t = (v_0 \cos \theta) t$ (Eq 1)

Step 2: Vertical Motion (Find Time)
Total displacement $\Delta y = 0$ .
$0 = (v_0 \sin \theta)t - \frac{1}{2}gt^2 \Rightarrow t = \frac{2v_0 \sin \theta}{g}$ (Eq 2)

Step 3: Substitute and Solve
$R = (v_0 \cos \theta)(\frac{2v_0 \sin \theta}{g}) = \frac{2v_0^2 \sin \theta \cos \theta}{g}$

Derivation 2: Maximum Height ( $H$ )
Find the peak height derived in terms of $v_0$ , $\theta$ , and $g$ .

Click to see Step-by-Step Derivation

The Key Concept:
At the very peak of the flight, the vertical velocity ( $v_y$ ) is zero.

Step 1: Choose the Right Equation
We don’t know time, so use the time-independent equation:
$v_y^2 = v_{0y}^2 + 2a_y \Delta y$

Step 2: Substitute Variables
Final vertical velocity $v_y = 0$
Initial vertical velocity $v_{0y} = v_0 \sin \theta$
Acceleration $a_y = -g$
Displacement $\Delta y = H$

Step 3: Solve for $H$
$0 = (v_0 \sin \theta)^2 - 2gH$
$2gH = v_0^2 \sin^2 \theta$
$H = \frac{v_0^2 \sin^2 \theta}{2g}$

Derivation 3: Equation of Trajectory (Path)
Prove that the path is a parabola by finding $y$ as a function of $x$ .

Click to see Step-by-Step Derivation

Goal: Eliminate time ( $t$ ) from the equations.

Step 1: Solve for $t$ using Horizontal Equation
$x = v_{0x} t \Rightarrow x = (v_0 \cos \theta) t$
$t = \frac{x}{v_0 \cos \theta}$ (Eq A)

Step 2: Plug $t$ into the Vertical Equation
$y = v_{0y}t - \frac{1}{2}gt^2$
$y = (v_0 \sin \theta)\left( \frac{x}{v_0 \cos \theta} \right) - \frac{1}{2}g \left( \frac{x}{v_0 \cos \theta} \right)^2$

Step 3: Simplify
Use the identity $\frac{\sin \theta}{\cos \theta} = \tan \theta$ :
$y = (\tan \theta)x - \left( \frac{g}{2v_0^2 \cos^2 \theta} \right) x^2$

Notice that this follows the form $y = Ax - Bx^2$ , which is mathematically a downward-opening parabola.

6. Don’t Lose Easy Points!

❌ The “Velocity at Peak” Trap

Mistake: Saying velocity is zero at the peak.

Truth: Only the vertical velocity ( $v_y$ ) is zero. The horizontal velocity ( $v_x$ ) is still there!

❌ The “Acceleration” Trap

Mistake: Saying acceleration is zero at the peak.

Truth: Acceleration is always $9.8 \, \text{m/s}^2$ downwards, even at the very top.

Ready for the next step?

Go to Unit 2: Forces »