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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 »

« Back to AP Physics Guide / Unit 1: Kinematics

AP Physics 1: Projectile Motion

Projectile motion describes the two-dimensional motion of an object launched into the air under the influence of gravity. Although the object’s path appears complex, every projectile problem can be solved by analyzing two simple motions occurring simultaneously: horizontal motion and vertical motion.

AP Exam Tip: Treat projectile motion as two independent one-dimensional problems. Solve the horizontal and vertical motions separately, then combine the results to fully describe the object’s motion.

Illustration of a projectile following a parabolic trajectory with horizontal and vertical motion components. — **Figure 1.1.** Projectile motion combines constant horizontal velocity with constant vertical acceleration due to gravity, producing a parabolic path.

Learning Objectives

By the end of this lesson, you will be able to:

Differentiate between scalars and vectors.
Resolve a vector into its horizontal and vertical components using trigonometry.
Explain why horizontal and vertical motions are independent.
Analyze projectile motion using separate x- and y-direction equations.
Calculate time of flight, maximum height, and horizontal range.
Solve AP Physics 1 free-response and multiple-choice projectile motion problems confidently.

Why this lesson matters
Projectile motion combines nearly every major idea learned in Unit 1—including vectors, kinematics equations, graph interpretation, and problem-solving strategies. Mastering this topic provides a strong foundation for later units involving forces, energy, and circular motion.

Scalars and Vectors

Before studying projectile motion, it is important to understand the difference between scalar and vector quantities. Every projectile problem involves vectors because both the magnitude and the direction of motion matter.

Comparison of scalar and vector quantities with everyday and physics examples. — **Figure 2.1.** Scalars have magnitude only, whereas vectors have both magnitude and direction.

Scalar Quantity	Vector Quantity
Has magnitude only.	Has both magnitude and direction.
Represented by a numerical value and unit.	Represented by a numerical value, unit, and direction (or an arrow).
Examples: distance, mass, time, temperature, energy, speed.	Examples: displacement, velocity, acceleration, force, momentum.

Why Are Vectors Important in Projectile Motion?

When an object is launched into the air, it does not move only upward or only forward. Instead, its motion occurs simultaneously in two perpendicular directions:

Horizontal (x-direction): Determines how far the projectile travels.
Vertical (y-direction): Determines how high the projectile rises and how long it remains in the air.

Since motion occurs in two directions at the same time, the initial velocity must be treated as a vector. In the next section, this vector will be resolved into horizontal and vertical components, allowing each direction to be analyzed independently.

Key Idea
A scalar answers “How much?”
A vector answers “How much and in which direction?”

Representing Vectors

In physics, vectors are represented by arrows. The length of the arrow represents the vector’s magnitude, while the arrowhead indicates its direction. Before solving projectile motion problems, it is important to understand how vectors are drawn and interpreted.

Illustration showing how vectors are represented by arrows with magnitude and direction. — **Figure 2.2.** A vector is represented by an arrow. The arrow’s length indicates magnitude, and the arrowhead shows direction.

Parts of a Vector

Part	Meaning
Magnitude	The numerical size or amount of the vector.
Direction	The way in which the vector points, usually measured as an angle.
Tail	The point where the vector begins.
Head	The arrowhead that indicates the vector’s direction.

Vector Notation

Vectors are commonly written using a bold letter (such as v) or with an arrow above the symbol ( $\vec{v}$ ). In AP Physics 1, vectors are usually identified by their magnitude and direction rather than using unit vector notation.

AP Exam Tip
Whenever a problem gives an initial velocity and a launch angle, think of the velocity as a single vector that will later be separated into horizontal and vertical components.

Looking Ahead
The next section shows how a vector can be resolved into two perpendicular components using trigonometry. This is the key step in solving every projectile motion problem.

Resolving Vectors into Components

A projectile is launched with a single initial velocity, but that velocity acts in two different directions simultaneously. To analyze the motion, the initial velocity vector is separated into two perpendicular components: a horizontal component and a vertical component. This process is called vector resolution or resolving a vector into components.

Horizontal and Vertical Components

Suppose an object is launched with an initial speed v at an angle θ above the horizontal.

Horizontal component

$v_x=v\cos\theta$

Vertical component

$v_y=v\sin\theta$

The horizontal component determines how fast the projectile moves across the ground, while the vertical component determines how high it rises and how long it remains in the air.

Why Cosine for the Horizontal Component?

The launch angle θ is measured from the horizontal axis. In the right triangle formed by the vector and its components, the horizontal component lies adjacent to the angle.

Adjacent side → Cosine
Opposite side → Sine

Therefore,

$v_x=v\cos\theta$
$v_y=v\sin\theta$

Memory Trick
If the launch angle is measured from the horizontal,

Horizontal → Cosine
Vertical → Sine

AP Exam Tip
Always resolve the initial velocity before substituting values into the kinematics equations. Solving directly with the original velocity usually leads to incorrect answers.

Common Mistake
Students often interchange sine and cosine. Remember that the correct component depends on where the angle is measured. In AP Physics 1, the launch angle is almost always measured from the horizontal axis.

Independence of Horizontal and Vertical Motion

The key to solving every projectile motion problem is understanding that the horizontal and vertical motions are independent. Although the object follows a single curved path, its motion can be analyzed as two separate one-dimensional motions occurring at the same time.

Gravity acts only in the vertical direction. Therefore, the horizontal motion is unaffected by gravity, while the vertical motion experiences a constant downward acceleration.

Illustration showing that horizontal and vertical motions of a projectile are independent. — **Figure 2.4.** Projectile motion is the combination of constant horizontal velocity and uniformly accelerated vertical motion. The two motions occur simultaneously but independently.

Horizontal Motion

In the absence of air resistance, no horizontal force acts on the projectile after it leaves the launcher. According to Newton’s First Law, the projectile continues moving with a constant horizontal velocity.

Horizontal acceleration: $a_x=0$
Horizontal velocity remains constant.
Horizontal displacement increases uniformly with time.

Vertical Motion

The vertical motion is affected only by gravity. The projectile accelerates downward throughout its flight, causing the vertical velocity to decrease while rising, become zero at the highest point, and increase downward during the descent.

Vertical acceleration: $a_y=-g$
Vertical velocity changes continuously.
Maximum height occurs when $v_y=0$ .

Comparing the Two Motions

Horizontal Motion	Vertical Motion
Acceleration $a_x=0$	Acceleration $a_y=-g$
Velocity remains constant	Velocity changes with time
Determines the horizontal range	Determines height and time of flight
Independent of gravity	Controlled entirely by gravity

AP Exam Tip
Always solve the horizontal and vertical motions separately. The only quantity shared by both motions is time. Once the time is known from one direction, it can immediately be used in the other.

Common Mistake
Many students believe that gravity slows the projectile in the horizontal direction. This is incorrect. Gravity acts only downward, so the horizontal velocity remains constant when air resistance is neglected.

Real-World Examples

A basketball shot follows a curved path because the horizontal and vertical motions occur simultaneously.
A cannonball continues moving forward while gravity pulls it downward.
Water leaving a fountain forms a projectile because it has both horizontal and vertical motion.

Key Takeaway
A projectile follows one curved trajectory, but it is always analyzed as two independent motions:

Horizontal motion → Constant velocity
Vertical motion → Constant acceleration due to gravity
Time connects both motions.

5. Projectile Motion Equations

Once the initial velocity has been resolved into horizontal and vertical components, projectile motion can be analyzed using two separate sets of kinematics equations. Since the horizontal and vertical motions are independent, each direction is solved separately before combining the results.

AP Exam Tip: Always begin by resolving the launch velocity into its components. Then use the vertical motion to determine the time whenever possible, and substitute that time into the horizontal motion equations.

Projectile motion equations showing horizontal motion, vertical motion, time of flight, maximum height, and horizontal range. — **Figure 2.5.** Projectile motion is solved using two independent sets of equations: constant velocity in the horizontal direction and constant acceleration in the vertical direction.

Horizontal Motion (x-direction)

Since no horizontal force acts on the projectile (neglecting air resistance), its horizontal acceleration is zero. Therefore, the horizontal velocity remains constant throughout the motion.

Horizontal Equations

$a_x=0$

$v_x=v_0\cos\theta$

$x=v_xt$

Vertical Motion (y-direction)

Gravity produces a constant downward acceleration throughout the flight. The vertical velocity continuously changes until the projectile reaches its maximum height, after which it increases downward.

Vertical Equations

$a_y=-g$

$v_y=v_0\sin\theta-gt$

$y=v_0\sin\theta\,t-\frac12gt^2$

Important Projectile Motion Formulas

For projectiles launched and landing at the same vertical level, several useful formulas can be derived directly from the kinematics equations.

Quantity	Formula	Meaning
Time of Flight	$T=\frac{2v_0\sin\theta}{g}$	Total time the projectile remains in the air.
Maximum Height	$H=\frac{v_0^2\sin^2\theta}{2g}$	Highest vertical position reached.
Horizontal Range	$R=\frac{v_0^2\sin2\theta}{g}$	Total horizontal distance traveled.

Choosing the Correct Equation

Every projectile problem becomes much easier after identifying the unknown quantity. Select the equation that contains only one unknown variable whenever possible.

General Strategy

Resolve the initial velocity into horizontal and vertical components.
Use the vertical equations to calculate the required time or height.
Substitute the time into the horizontal equation to determine the range or horizontal displacement.
Combine both directions to describe the complete motion.

Common Mistake
Do not mix horizontal and vertical equations. The horizontal motion has zero acceleration, whereas the vertical motion always has an acceleration of $-g$ . Keep the two directions separate throughout the calculation.

Memory Trick

Horizontal → Constant Velocity
Vertical → Constant Acceleration
Time links both motions.

Key Takeaway
Projectile motion is not a new type of motion. It is simply the combination of two familiar one-dimensional motions:

Horizontal motion with constant velocity.
Vertical motion with constant acceleration due to gravity.
Both motions occur simultaneously and are connected only through time.

8. Time of Flight, Maximum Height, and Horizontal Range

Once the initial velocity has been separated into horizontal and vertical components, three important quantities can be calculated for every projectile:

Quantity	Meaning	Symbol
Time of Flight	Total time the projectile remains in the air.	T
Maximum Height	Highest vertical position reached.	H
Horizontal Range	Total horizontal distance traveled.	R

These equations apply when the projectile is launched and lands at the same vertical height, which is the most common situation encountered in AP Physics 1.

Time of Flight

The projectile rises and falls under constant acceleration due to gravity. Since the upward journey and downward journey take equal amounts of time (for equal launch and landing heights), the total time of flight is:

T = 2v₀ sinθ / g

Maximum Height

At the highest point of the trajectory, the vertical velocity becomes zero. Applying the kinematic equation gives:

H = (v₀² sin²θ)/(2g)

Horizontal Range

The range equals the constant horizontal velocity multiplied by the total time of flight. Combining these relationships produces:

R = (v₀² sin2θ)/g

AP Exam Tip

Always verify whether the projectile lands at the same height before using these formulas.
If launch and landing heights differ, use the kinematic equations instead of memorized formulas.
Remember that sin(2θ) appears only in the horizontal range equation.

Common Mistakes

Using the range equation when the projectile lands at a different height.
Replacing sin²θ with sin(2θ).
Forgetting that the vertical velocity is zero only at the highest point.
Using horizontal velocity in the maximum height calculation.

Key Takeaway

Time of Flight depends on the vertical component of velocity.
Maximum Height depends only on vertical motion.
Horizontal Range depends on both horizontal and vertical motion.
All three equations are derived from the same kinematic principles.

9. Solving Projectile Motion Problems

Most AP Physics 1 projectile questions can be solved using the same systematic approach. Rather than memorizing many formulas, focus on identifying the known quantities and solving the horizontal and vertical motions separately.

AP Exam Tip: Whenever a projectile problem appears, draw a diagram first. Label the initial velocity, launch angle, horizontal and vertical components, and the direction of gravity before writing any equations.

Flowchart showing the step-by-step strategy for solving projectile motion problems in AP Physics. — **Figure 2.6.** A systematic approach to solving projectile motion problems reduces mistakes and improves accuracy on the AP Physics exam.

Step 1 — Draw the Situation

Sketch the projectile’s path and identify:

Launch angle (θ)
Initial speed (v₀)
Horizontal direction (x-axis)
Vertical direction (y-axis)
Acceleration due to gravity (g)

Step 2 — Resolve the Initial Velocity

Convert the initial velocity into horizontal and vertical components.

$v_x=v_0\cos\theta$

$v_y=v_0\sin\theta$

Step 3 — Solve the Vertical Motion

The vertical motion determines the important time values because gravity acts only in this direction.

Time of flight
Maximum height
Vertical position
Final vertical velocity

Step 4 — Solve the Horizontal Motion

After finding the required time, substitute it into the horizontal equation.

$x=v_xt$

This gives the horizontal displacement or range.

Step 5 — Check the Answer

Include SI units.
Check significant figures.
Verify the direction.
Ask whether the answer is physically reasonable.

Quick Checklist

Draw the diagram.
Resolve vectors.
Solve vertical motion.
Find the required time.
Solve horizontal motion.
Write the answer with units.

Common Mistakes

Using the total velocity instead of its components.
Mixing horizontal and vertical equations.
Using +g instead of −g.
Ignoring units.
Applying memorized formulas when launch and landing heights are different.

Key Takeaway
Every projectile motion problem becomes much simpler after separating the motion into two independent one-dimensional motions. Nearly every AP Physics question follows this same sequence of steps.

10. Worked Example: Finding the Maximum Height

Now let’s apply everything learned so far by solving a complete projectile motion problem step by step. The goal is not simply to obtain the answer, but to understand the reasoning behind every calculation.

AP Exam Tip: Write the known values first. Most mistakes occur because students substitute numbers before identifying the correct equation.

Worked example showing how to calculate the maximum height of a projectile launched at an angle. — **Figure 2.7.** A complete AP Physics 1 example demonstrating how to calculate the maximum height of a projectile.

Problem

A ball is launched with an initial speed of 20 m/s at an angle of 45° above the horizontal.

Find the maximum height reached by the ball.

Step 1 — Identify the Known Quantities

Initial speed: $v_0=20\text{ m/s}$
Launch angle: $\theta=45^\circ$
Acceleration: $g=9.8\text{ m/s}^2$
At maximum height: $v_y=0$

Step 2 — Find the Initial Vertical Velocity

Resolve the launch velocity into its vertical component.

$v_y=v_0\sin\theta$

$v_y=20\sin45^\circ$

$v_y=20(0.707)=14.14\text{ m/s}$

Step 3 — Apply the Vertical Motion Equation

Use the equation relating velocity and displacement.

$v_y^2=u_y^2-2gH$

Since the vertical velocity at the highest point is zero,

$0=(14.14)^2-2(9.8)H$

$H=\frac{200}{19.6}$

$H=10.2\text{ m}$

Final Answer

Maximum Height = 10.2 m

Common Mistake

Using the total initial velocity (20 m/s) instead of the vertical component (14.14 m/s).
Using $g=+9.8$ instead of treating gravity as downward in the equation.
Forgetting that the vertical velocity becomes zero only at the highest point.

Key Takeaway
Every maximum-height problem begins by resolving the launch velocity into vertical and horizontal components. Only the vertical motion is needed to calculate the highest point.

11. Worked Example: Time of Flight and Horizontal Range

In this example, the projectile is launched and lands at the same vertical height. This allows the standard projectile motion formulas to be used directly. The example demonstrates how to determine both the total time the projectile remains in the air and the horizontal distance it travels.

AP Exam Tip: Whenever a projectile lands at the same height from which it was launched, the formulas for time of flight and horizontal range can be applied directly.

Worked example showing the calculation of time of flight and horizontal range of a projectile. — **Figure 2.8.** A complete AP Physics 1 example demonstrating how to calculate both the time of flight and the horizontal range of a projectile.

Problem

A football is kicked with an initial speed of 25 m/s at an angle of 37° above the horizontal.

Find:

The total time of flight.
The horizontal range.

Step 1 — Identify the Known Quantities

Initial speed: $v_0=25\text{ m/s}$
Launch angle: $\theta=37^\circ$
Acceleration due to gravity: $g=9.8\text{ m/s}^2$
Launch height = Landing height

Step 2 — Calculate the Time of Flight

Use the standard equation:

$T=\frac{2v_0\sin\theta}{g}$

$T=\frac{2(25)\sin37^\circ}{9.8}$

$T=\frac{50(0.602)}{9.8}$

$T=\frac{30.1}{9.8}$

$T=3.07\text{ s}$

Step 3 — Calculate the Horizontal Range

Use the range equation:

$R=\frac{v_0^2\sin2\theta}{g}$

$R=\frac{25^2\sin74^\circ}{9.8}$

$R=\frac{625(0.961)}{9.8}$

$R=\frac{600.6}{9.8}$

$R=61.3\text{ m}$

Final Answers

Time of Flight = 3.07 s

Horizontal Range = 61.3 m

Solution Strategy

Check whether the projectile lands at the same height.
Calculate the time of flight.
Use the range equation or multiply horizontal velocity by time.
Write the final answer with SI units.

Common Mistakes

Using the range equation when the landing height is different.
Forgetting that the angle inside the range equation is 2θ.
Using degrees incorrectly in the calculator.
Omitting units in the final answer.

Key Takeaway
For projectiles launched and landing at the same height, the time of flight depends on the vertical motion, while the horizontal range depends on both the horizontal velocity and the total flight time.

12. Practice Problems (Easy → Medium → AP Level)

Now it’s time to apply the concepts learned throughout this lesson. These practice problems gradually increase in difficulty, beginning with straightforward calculations and progressing to AP Physics 1 style conceptual and multi-step questions.

AP Physics 1 Projectile Motion practice problems ranging from easy to AP exam level with conceptual questions and problem-solving strategy. — **Figure 12.1.** Practice problems arranged from easy to AP Physics 1 exam level, designed to strengthen understanding of projectile motion, vector components, and multi-step problem solving.

AP Exam Tip: Always begin by drawing the projectile, identifying the horizontal and vertical components, and listing the known quantities before selecting an equation.

Level 1 — Easy

A ball is launched horizontally from a table with a speed of 8 m/s. What is its horizontal velocity after 2 seconds?
A projectile is launched vertically upward with an initial speed of 20 m/s. Calculate its maximum height.
An object is projected at 15 m/s making an angle of 30°. Determine its horizontal and vertical velocity components.
A projectile remains in the air for 4 seconds. What is its maximum height?

Level 2 — Medium

A football is kicked at 22 m/s and 40°. Calculate the total time of flight.
A projectile is launched at 28 m/s making an angle of 45°. Find the horizontal range.
A cannonball is fired at 35 m/s at an angle of 50°. Determine the maximum height.
A projectile has horizontal and vertical velocity components of 18 m/s and 24 m/s. Calculate its launch speed and launch angle.

Level 3 — AP Physics 1 Challenge

A ball is kicked with an initial speed of 30 m/s at an angle of 40°. Determine:
- Total time of flight
- Maximum height
- Horizontal range
A projectile lands 82 m away after 3.8 s. Determine its horizontal velocity.
A projectile reaches a maximum height of 18 m. Determine its initial vertical velocity.
Two projectiles are launched with the same speed but at angles of 30° and 60°. Which one travels farther? Explain your reasoning.

Concept Check

Why does gravity affect only the vertical motion of a projectile?
At the highest point of a projectile, which velocity component becomes zero?
Does the horizontal velocity ever become zero during projectile motion? Explain.
Which quantity remains constant throughout projectile motion (neglecting air resistance)?
Why is projectile motion treated as two independent one-dimensional motions?

Challenge Yourself
Solve every problem without looking at the worked examples. If an answer seems unreasonable, check the units, redraw the diagram, and verify whether each equation applies to horizontal motion or vertical motion.

Learning Goal
After completing these practice problems, solving AP Physics 1 projectile motion questions should become a systematic process:

Draw the diagram.
Resolve the velocity into components.
Analyze horizontal and vertical motion separately.
Select the appropriate equation.
Check units and reasonableness of the final answer.

13. Common AP Physics 1 Mistakes

Many errors in projectile motion are caused by using the correct equations in the wrong direction or confusing horizontal and vertical quantities. Learning to recognize these common mistakes will improve accuracy and save valuable marks on the AP Physics 1 exam.

Infographic showing the most common AP Physics 1 projectile motion mistakes, including mixing horizontal and vertical motion, gravity misconceptions, incorrect trigonometric functions, forgetting units, and assuming velocity is zero at the highest point. — **Figure 13.1.** Common mistakes made in AP Physics 1 projectile motion problems. Keeping horizontal and vertical motions separate, using the correct trigonometric functions, and checking units can significantly improve accuracy on both multiple-choice and free-response questions.

AP Exam Tip: Before solving any projectile problem, label every known quantity as either horizontal (x) or vertical (y). Mixing these variables is the most common source of mistakes.

Mistake 1 — Mixing Horizontal and Vertical Motion

Horizontal equations should only contain horizontal quantities, while vertical equations should only contain vertical quantities. Never substitute a vertical velocity into a horizontal equation.

Mistake 2 — Forgetting that Gravity Acts Only Vertically

Gravity changes only the vertical velocity. The horizontal velocity remains constant throughout projectile motion when air resistance is neglected.

Mistake 3 — Using the Wrong Trigonometric Function

When the launch angle is measured from the horizontal,

Horizontal component = v₀ cos θ
Vertical component = v₀ sin θ

Always identify the adjacent and opposite sides before selecting sine or cosine.

Mistake 4 — Assuming Velocity is Zero at the Highest Point

Only the vertical component of velocity becomes zero at the highest point. The projectile continues moving horizontally with constant velocity.

Mistake 5 — Ignoring Units

Always write answers with SI units:

Displacement → m
Velocity → m/s
Acceleration → m/s²
Time → s

Quick Checklist Before Submitting an Answer

✓ Resolved the initial velocity into components.
✓ Used separate equations for x- and y-motion.
✓ Used the correct trigonometric function.
✓ Included the effect of gravity only in the vertical direction.
✓ Checked units and significant figures.
✓ Verified that the answer is physically reasonable.

Remember
Most AP Physics projectile motion questions become much easier after drawing a diagram, resolving the initial velocity into components, and treating horizontal and vertical motions independently.

14. Concept Check / Quick Quiz

Before moving on to AP-style multiple-choice questions, take a few minutes to check your understanding of the core concepts covered in this lesson. These questions focus on reasoning rather than lengthy calculations.

AP Exam Tip: Many AP Physics 1 questions test conceptual understanding rather than computation. Always identify the physical principle before reaching for an equation.

Quick Quiz

A projectile reaches its highest point. Which velocity component is zero?
Which velocity component remains constant throughout projectile motion (neglecting air resistance)?
Why is projectile motion considered two independent one-dimensional motions?
A ball is launched at an angle of 45°. Which trigonometric function is used to determine the horizontal component of the initial velocity?
If the launch speed is doubled while the launch angle remains the same, what happens to the horizontal velocity component?
At every point during projectile motion, what is the direction of gravitational acceleration?
Which of the following quantities remains constant during projectile motion?
- Horizontal velocity
- Vertical velocity
- Acceleration due to gravity
- Both A and C
True or False: At the highest point of the trajectory, the projectile is momentarily at rest.
A projectile is launched horizontally from a cliff. What is its initial vertical velocity?
Which quantity determines how far a projectile travels horizontally?

Concept Check infographic for Projectile Motion showing ten conceptual AP Physics 1 questions, exam tips, self-evaluation guide, and key reminders. — **Figure 14.1.** Concept Check for Projectile Motion. Test conceptual understanding of velocity components, vector resolution, gravity, independence of motion, and key projectile motion principles before attempting AP-style questions.

Answers

Vertical velocity (v_y)
Horizontal velocity (v_x)
Gravity acts only in the vertical direction, so horizontal and vertical motions can be analyzed separately.
Cosine
It doubles.
Vertically downward.
Both A and C
False
Zero
The horizontal velocity and the total time of flight.

Self-Evaluation

9–10 correct: Excellent! Ready for AP-style questions.
7–8 correct: Good understanding. Review the vector components and projectile equations.
Below 7: Revisit the worked examples before attempting the AP-level questions.

15. Multiple Choice Questions (AP Physics 1 Style)

The following questions are modeled after the style and reasoning expected on the AP Physics 1 examination. Some questions require calculations, while others test conceptual understanding. Attempt every question before checking the answer key.

AP Exam Tip: Read every answer choice carefully. Many incorrect options represent common misconceptions about projectile motion, vectors, or kinematics.

Questions

A projectile reaches its highest point. Which quantity is zero?
1. Horizontal velocity
2. Vertical velocity
3. Acceleration
4. Horizontal acceleration
Ignoring air resistance, which quantity remains constant throughout projectile motion?
1. Vertical velocity
2. Resultant velocity
3. Horizontal velocity
4. Height
A ball is launched with speed v at an angle θ above the horizontal. Which expression gives the horizontal component of the initial velocity?
1. v sin θ
2. v cos θ
3. v tan θ
4. v/sec θ
A projectile is launched horizontally from a cliff. Its initial vertical velocity is
1. g
2. −g
3. 0
4. Depends on height
At every point during projectile motion, the acceleration is
1. Horizontal
2. Along the direction of motion
3. Vertically downward
4. Zero
A projectile is launched at 20 m/s and 30°. The horizontal component of the initial velocity is approximately
1. 10 m/s
2. 17.3 m/s
3. 20 m/s
4. 15 m/s
Which launch angle gives the greatest horizontal range when launch and landing occur at the same height?
1. 30°
2. 45°
3. 60°
4. 75°
A projectile has reached its maximum height. Which statement is correct?
1. Both velocity components are zero.
2. The projectile stops moving.
3. Vertical velocity is zero but acceleration remains downward.
4. Acceleration becomes zero.
If the launch speed doubles while the angle remains unchanged, the horizontal component
1. Halves
2. Remains unchanged
3. Doubles
4. Quadruples
Which equation is used to determine the maximum height?
1. v = u + at
2. v² = u² + 2as
3. s = ut
4. P = mv
Neglecting air resistance, the path of a projectile is
1. Linear
2. Circular
3. Parabolic
4. Elliptical
Which factor affects the time of flight?
1. Horizontal velocity only
2. Vertical component of initial velocity
3. Mass only
4. Horizontal range

Answer Key

Q	Ans	Q	Ans	Q	Ans
1	B	5	C	9	C
2	C	6	B	10	B
3	B	7	B	11	C
4	C	8	C	12	B

Ready for the next challenge?
The next section contains a complete AP Physics 1 Free-Response Question (FRQ) with a detailed, step-by-step solution similar to those found on the actual AP exam.

16. AP Physics 1 Free Response Question (FRQ)

The following free-response question is modeled after the style of the College Board AP Physics 1 examination. Show all work, include diagrams where appropriate, justify each step, and clearly state the final answer with proper units.

AP Exam Tip: On the AP exam, partial credit is awarded for correct reasoning even if the final numerical answer is incorrect. Label vectors, write equations before substituting numbers, and include units throughout your solution.

Question

A soccer ball is kicked from ground level with an initial speed of 20.0 m/s at an angle of 40° above the horizontal. Neglect air resistance.

Calculate:

The horizontal and vertical components of the initial velocity.
The total time of flight.
The maximum height reached.
The horizontal range of the projectile.

Take g = 9.8 m/s².

Complete Solution

Step 1. Resolve the initial velocity

v_x = v cos θ = 20 cos 40° ≈ 15.3 m/s

v_y = v sin θ = 20 sin 40° ≈ 12.9 m/s

Step 2. Time of flight

T = 2v_y/g = 2(12.9)/9.8 ≈ 2.63 s

Step 3. Maximum height

H = v_y² / 2g = (12.9)² / (2×9.8) ≈ 8.5 m

Step 4. Horizontal range

R = v_xT = 15.3 × 2.63 ≈ 40.2 m

Final Answers

Horizontal velocity: 15.3 m/s
Vertical velocity: 12.9 m/s
Time of flight: 2.63 s
Maximum height: 8.5 m
Horizontal range: 40.2 m

Scoring Guide (AP Style)

1 point — Correct vector components
1 point — Correct time of flight
1 point — Correct maximum height
1 point — Correct horizontal range
1 point — Appropriate equations, reasoning, and units

17. Real-World Applications of Projectile Motion

Projectile motion is much more than a classroom topic. Engineers, athletes, scientists, and military professionals use the same principles to predict the motion of objects traveling through the air. Understanding projectile motion helps explain how objects move under the influence of gravity and allows accurate predictions of where they will land.

AP Exam Tip: Whenever a real-world problem involves an object moving through the air without propulsion after launch, begin by separating the motion into independent horizontal and vertical components.

Real-world applications of projectile motion including sports, engineering, firefighting, military, space science, and entertainment. — **Figure 17.1.** Projectile motion is applied in sports, engineering, emergency services, space exploration, military science, and entertainment to predict trajectories accurately.

Common Applications

Field	Application of Projectile Motion
Sports	Basketball shots, football passes, cricket throws, golf drives, javelin, and long jump analysis.
Engineering	Designing water fountains, ballistics software, robotic launch systems, and safety simulations.
Firefighting	Selecting the correct angle and pressure for water jets to reach elevated locations.
Military	Predicting artillery shell trajectories and targeting systems.
Space Science	Planning rocket launches, satellite deployment, and spacecraft landing trajectories.
Entertainment	Creating realistic animations and physics-based simulations in video games and films.

Why Projectile Motion Matters

Predicts where an object will land.
Improves accuracy in sports and engineering.
Supports safer designs and better planning.
Forms the foundation for advanced mechanics and orbital motion.
Demonstrates how mathematics models real physical systems.

Key Idea
Although real objects often experience air resistance, the ideal projectile model provides an excellent approximation for many practical situations and serves as the basis for more advanced motion analysis.

18. Summary / Key Takeaways

Projectile motion becomes much easier when the motion is separated into its horizontal and vertical components. The key is to treat these two motions independently while remembering that they occur simultaneously.

The Big Idea

A projectile has: constant horizontal velocity and constant vertical acceleration due to gravity. Together, these produce a parabolic trajectory.

AP Physics 1 projectile motion summary showing key concepts, horizontal and vertical motion, projectile trajectory, and problem-solving strategy. — **Figure 18.1.** Projectile Motion Summary and Key Takeaways — a visual review of horizontal and vertical motion, vector components, the projectile trajectory, and the AP problem-solving strategy.

Key Concepts to Remember

Projectile motion is two-dimensional motion that can be analyzed as independent horizontal and vertical motions.
The initial velocity can be resolved into two components: v_x and v_y.
The horizontal acceleration is zero: a_x = 0.
The vertical acceleration is constant and directed downward: a_y = −g.
Horizontal velocity remains constant when air resistance is neglected.
Vertical velocity changes continuously because of gravity.
At the highest point, the vertical velocity is zero: v_y = 0.
The acceleration is still directed downward at the highest point.
For launch and landing at the same height, the time of ascent equals the time of descent.
The horizontal range depends on horizontal velocity and total time of flight.

The Problem-Solving Strategy

Draw the situation. Identify the launch angle, initial velocity, height, and range.
Choose the x- and y-axes.
Resolve the initial velocity into horizontal and vertical components.
Analyze horizontal motion using constant velocity.
Analyze vertical motion using constant acceleration due to gravity.
Use the appropriate kinematics equation.
Combine the horizontal and vertical results to obtain the required quantity.
Check the answer for reasonable magnitude, direction, and units.

AP Physics 1 Exam Reminder

Projectile motion problems often look complicated because the object moves in two dimensions. The mathematics becomes much simpler when the motion is separated into independent x- and y-direction problems.

Ready for the next step?
The next section provides a compact Projectile Motion Formula Sheet containing the equations and relationships needed for quick AP Physics 1 revision.

19. Projectile Motion Formula Sheet

Use this formula sheet for quick revision before attempting AP Physics 1 projectile-motion questions. The equations are organized by horizontal motion, vertical motion, and common projectile results.

AP Exam Tip: Do not memorize equations without understanding when they apply. First identify the direction of motion, known quantities, and acceleration.

AP Physics 1 projectile motion formula sheet showing velocity components, horizontal and vertical kinematics, time of flight, maximum height, and range. — **Figure 19.1.** Projectile Motion Formula Sheet for AP Physics 1 quick revision.

1. Initial Velocity Components

Horizontal: v_0x = v₀ cos θ

Vertical: v_0y = v₀ sin θ

2. Horizontal Motion

a_x = 0

v_x = v_0x = constant

x = x₀ + v_0xt

3. Vertical Motion

a_y = −g

v_y = v_0y − gt

y = y₀ + v_0yt − ½gt²

v_y² = v_0y² − 2g(y − y₀)

4. Common Projectile Results

Time to maximum height: t_up = v_0y/g

Maximum height above launch point: H = v_0y²/(2g)

Total time of flight, same launch and landing height: T = 2v_0y/g

Horizontal range, same launch and landing height: R = v_0xT

5. Direct Formulas for Launch Speed and Angle

Maximum height: H = v₀²sin²θ/(2g)

Time of flight: T = 2v₀sin θ/g

Range: R = v₀²sin(2θ)/g

Important Condition

The commonly used formulas for total time of flight and horizontal range assume that the projectile is launched and lands at the same vertical height, with air resistance neglected.

Quick Memory Guide

Horizontal → cosine → constant velocity
Vertical → sine → acceleration −g
Highest point → v_y = 0
Gravity → always downward
Range → horizontal velocity × time

20. Frequently Asked Questions About Projectile Motion

Projectile motion can seem difficult at first because an object is moving horizontally and vertically at the same time. The questions below address the most common conceptual and calculation-related questions about projectile motion in AP Physics 1.

Quick Reminder: For ideal projectile motion, air resistance is neglected and gravity is the only acceleration acting on the projectile.

Frequently asked questions about projectile motion for AP Physics 1, including horizontal and vertical motion, velocity, acceleration, maximum height, time of flight, and range. — **Figure 20.1.** Frequently Asked Questions about Projectile Motion — a visual review of key AP Physics 1 concepts, equations, common mistakes, and quick-revision strategies.

1. What is projectile motion?

Projectile motion is the two-dimensional motion of an object launched into the air that moves under the influence of gravity, assuming air resistance is neglected. The motion can be separated into horizontal and vertical components.

2. Why is projectile motion divided into horizontal and vertical motion?

The horizontal and vertical motions are independent of each other. Gravity produces vertical acceleration, while there is no horizontal acceleration in the ideal projectile-motion model. Both motions occur during the same time interval.

3. Is horizontal velocity constant in projectile motion?

Yes. When air resistance is neglected, horizontal acceleration is zero, so the horizontal velocity remains constant throughout the motion.

4. What happens to vertical velocity during projectile motion?

Vertical velocity changes because the projectile experiences a constant downward acceleration due to gravity. During upward motion, the vertical velocity decreases until it becomes zero at maximum height. During downward motion, the vertical velocity increases in the downward direction.

5. Is the velocity zero at the highest point of a projectile’s path?

No. Only the vertical component of velocity is zero at the highest point. If the projectile has horizontal velocity, it continues moving horizontally. Therefore, the projectile is not generally at rest at maximum height.

6. What is the acceleration of a projectile at its highest point?

The acceleration is still directed downward and has magnitude g. Reaching maximum height does not make the acceleration zero.

7. How are the initial velocity components calculated?

When the launch angle θ is measured above the horizontal, the initial velocity is resolved using:

v_0x = v₀ cos θ v_0y = v₀ sin θ

8. What is the acceleration of a projectile?

For ideal projectile motion near Earth’s surface, the horizontal acceleration is zero and the vertical acceleration is approximately 9.8 m/s² downward.

9. How is the maximum height of a projectile calculated?

For a projectile launched and landing at the same vertical level, the maximum height above the launch point can be found from the initial vertical velocity:

H = v_0y² / 2g

10. How is the time of flight calculated?

When the projectile lands at the same height from which it was launched, the total time of flight is:

T = 2v_0y / g

For launch and landing at different heights, the vertical kinematics equation must be used instead of this same-height formula.

11. What is the horizontal range of a projectile?

The horizontal range is the horizontal distance traveled by the projectile from its launch point to its landing point. It can be calculated using:

R = v_0xT

For launch and landing at the same height, the range can also be written as:

R = v₀² sin(2θ) / g

12. What launch angle gives the maximum range?

For a projectile launched and landing at the same height, with a fixed initial speed and negligible air resistance, the maximum theoretical horizontal range occurs at a launch angle of 45°.

13. Does mass affect projectile motion?

In the ideal projectile-motion model, mass does not affect the acceleration due to gravity. Therefore, two objects launched under the same conditions have the same ideal gravitational acceleration, provided air resistance is neglected.

14. What is the most common mistake in projectile-motion problems?

A common mistake is treating the entire velocity as either horizontal or vertical. The initial velocity must first be resolved into its horizontal and vertical components. Another common mistake is assuming that acceleration becomes zero at maximum height.

AP Physics 1 Key Reminder

Projectile motion is easiest to solve when the problem is separated into two parts:

x-direction: constant velocity, a_x = 0
y-direction: constant acceleration, a_y = −g

Still reviewing projectile motion?

Return to the Formula Sheet for quick revision, then test understanding with the Concept Check, AP-style MCQs, and FRQ before moving to the next lesson.

Continue Your AP Physics Journey

Congratulations on completing the Projectile Motion lesson! You should now be able to analyze two-dimensional motion by separating horizontal and vertical components, determine flight time, maximum height, horizontal range, and solve AP Physics projectile-motion questions. The next step is to move from kinematics to dynamics by learning how forces affect the motion of objects through Newton’s Laws and Free-Body Diagrams.

← Previous Lesson

AP Physics 1: Motion Graphs

Review position–time and velocity–time graphs, including slope, acceleration, displacement, graph interpretation, worked examples, and AP-style practice questions.

Review Motion Graphs

Next Lesson →

AP Physics 1: Newton’s Laws & Free Body Diagrams

Learn how forces cause changes in motion using Newton’s First, Second, and Third Laws. Explore net force, inertia, acceleration, mass, weight, and free-body diagrams with AP-style examples and practice problems.

Start Newton’s Laws & FBDs

Unit 1 Progress

✅ Kinematics Equations
✅ Motion Graphs
✅ Projectile Motion

Next Unit

Unit 2: Dynamics begins with Newton’s Laws & Free-Body Diagrams. The next lesson explains how forces cause changes in motion and introduces net force, inertia, acceleration, mass, weight, and free-body diagrams.

Start Unit 2: Newton’s Laws & FBDs