AP · Projectile Motion · 14 min read · Updated 2026-05-10

Projectile Motion — AP Physics 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: Independence of horizontal and vertical motion, kinematic equations for constant acceleration projectile motion, maximum height, time of flight, range calculations, and analysis of projectiles launched from/landing at different elevations.

You should already know: Kinematic equations for constant acceleration in one dimension. Vector resolution into perpendicular components. Acceleration due to gravity near Earth's surface.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 1 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Projectile Motion?

Projectile motion describes the motion of an object (called a projectile) that is launched into the air, and after launch, experiences only the constant downward force of gravity (air resistance is explicitly ignored per AP Physics 1 conventions). It is sometimes called ballistic motion in other sources, but the AP exam uniformly uses the term projectile motion. Per the AP Physics 1 Course and Exam Description (CED), projectile motion is part of Unit 1: Kinematics, which contributes 10–16% of the total exam score. Projectile motion questions appear in both multiple-choice (MCQ) and free-response (FRQ) sections: it is common as a standalone MCQ, or as the opening part of a multi-concept FRQ that connects projectile motion to forces, energy, or momentum later in the problem. We use standard AP notation throughout this guide: $v_{0}$ = initial speed, $θ_{0}$ = launch angle above the horizontal, $a_{x} = 0$ , $a_{y} = - g$ where $g = 9.8 m/s^{2}$ , origin placed at the launch point unless stated otherwise, positive y = upward, positive x = horizontal in the direction of initial motion. Adopting this consistent convention will eliminate most sign errors.

2. Independence of Perpendicular Motions

The core foundational principle of projectile motion is that horizontal and vertical motions are completely independent of one another, connected only by the shared time of flight. This is true because acceleration from gravity acts exclusively in the vertical direction, so there is no acceleration in the horizontal direction (with air resistance ignored). This means we can split any 2D projectile problem into two separate 1D constant acceleration problems, which we already know how to solve. First, we resolve the initial velocity vector into x and y components: $v_{0 x} = v_{0} cos θ_{0} v_{0 y} = v_{0} sin θ_{0}$ For the x-direction: since $a_{x} = 0$ , horizontal velocity is constant, so the kinematic equation simplifies to $x = x_{0} + v_{0 x} t$ . For the y-direction: acceleration is constant at $a_{y} = - g$ , so we use the full set of 1D kinematic equations: $v_{y} = v_{0 y} - g t$ , $y = y_{0} + v_{0 y} t - \frac{1}{2} g t^{2}$ , and $v_{y}^{2} = v_{0 y}^{2} - 2 g (y - y_{0})$ . This framework works for every projectile problem, regardless of launch angle or elevation.

Worked Example

A projectile is launched with initial speed 20 m/s at 30° above the horizontal. What are the x and y positions of the projectile after 1.0 second, relative to the launch point?

Set origin at the launch point, so $x_{0} = 0$ , $y_{0} = 0$ , and $t = 1.0 s$ .
Resolve initial velocity components: $v_{0 x} = 20 cos 3 0^{\circ} = 103 \approx 17.3 m/s$ , $v_{0 y} = 20 sin 3 0^{\circ} = 10 m/s$ .
Calculate x position: $x = v_{0 x} t = (17.3) (1.0) = 17.3 m \approx 17 m$ .
Calculate y position: $y = v_{0 y} t - \frac{1}{2} g t^{2} = (10) (1) - (4.9) (1^{2}) = 5.1 m$ .
Final result: After 1 second, the projectile is 17 m horizontally and 5.1 m vertically above the launch point.

Exam tip: Always draw and label your coordinate system explicitly on your paper for every projectile problem, marking which direction is positive y. This eliminates 90% of common sign errors on the AP exam.

3. Projectiles Launched From Level Ground

When a projectile is launched and lands at the same vertical elevation ( $y = y_{0}$ when it lands), we can derive simplified expressions for time of flight, maximum height, and range that speed up calculations for multiple-choice questions. Starting with time of flight: set $y - y_{0} = 0$ and substitute into the y-displacement equation: $0 = v_{0 y} t - \frac{1}{2} g t^{2} = t (v_{0 y} - \frac{1}{2} g t)$ The two solutions are $t = 0$ (the moment of launch) and the total time of flight: $t_{t o t a l} = \frac{2 v _{0 y}}{g} = \frac{2 v _{0} s i n θ _{0}}{g}$ . Maximum height occurs when vertical velocity $v_{y} = 0$ , which happens at $t_{ma x} = \frac{v _{0} s i n θ _{0}}{g}$ (half the total time, as expected for symmetric motion). Substituting back gives maximum height: $H = \frac{v _{0}^{2} s i n ^{2} θ _{0}}{2 g}$ . Range (total horizontal distance traveled) is $R = v_{0 x} t_{t o t a l}$ , which simplifies to: $R = \frac{v _{0}^{2} sin 2 θ _{0}}{g}$ From this formula, we get the key result that maximum range for level ground launch occurs at $θ_{0} = 4 5^{\circ}$ , since $sin 2 θ$ reaches its maximum value of 1 when $2 θ = 9 0^{\circ}$ .

Worked Example

A soccer player kicks a ball from ground level with an initial speed of 22 m/s at 35° above the horizontal. What is the maximum height reached by the ball, and what is its total range?

Confirm launch and landing are at the same elevation, so simplified formulas apply.
Calculate maximum height: $H = \frac{v _{0}^{2} s i n ^{2} θ _{0}}{2 g} = \frac{( 2 2 ^{2} ) ( s i n 3 5 ^{\circ} ) ^{2}}{2 ( 9.8 )} = \frac{484 ( 0.5736 ) ^{2}}{19.6} \approx \frac{484 ( 0.329 )}{19.6} \approx 8.1 m$ .
Calculate range: $R = \frac{v _{0}^{2} s i n 2 θ _{0}}{g} = \frac{484 ( s i n 7 0 ^{\circ} )}{9.8} \approx \frac{484 ( 0.940 )}{9.8} \approx 46 m$ .
Final result: Maximum height ≈ 8.1 m, total range ≈ 46 m.

Exam tip: The simplified range formula $R = \frac{v _{0}^{2} s i n 2 θ}{g}$ ONLY works for level ground (equal launch and landing elevation). Never use it for projectiles launched from cliffs or onto hills unless you confirm the elevations are equal.

4. Projectiles Launched From Uneven Elevation

Most non-routine projectile problems on the AP exam involve launch and landing at different elevations, such as throwing a ball off a building or launching a cannon at a target on a hill. For these problems, the simplified level-ground formulas do not apply, so we must return to the core kinematic equations. The standard solution method is: 1) write the y-displacement equation with the known final y position, 2) solve the resulting quadratic equation for time, 3) discard any non-physical negative time solution, 4) use the resulting valid time to solve for unknown horizontal distance or velocity. This method works for any projectile problem, regardless of elevation difference.

Worked Example

A cannon sits on top of a 45 m tall cliff. It fires a cannonball with initial speed 85 m/s at 25° above the horizontal, toward the flat ground below the cliff. What is the horizontal distance from the base of the cliff to where the cannonball lands?

Set origin at the cannon, so $y_{0} = 0$ , final $y = - 45 m$ (positive y = upward). Resolve initial velocity: $v_{0 x} = 85 cos 2 5^{\circ} \approx 77.0 m/s$ , $v_{0 y} = 85 sin 2 5^{\circ} \approx 36.0 m/s$ .
Write the y-displacement equation: $y = v_{0 y} t - 4.9 t^{2}$ , so $- 45 = 36 t - 4.9 t^{2}$ . Rearrange to standard quadratic form: $4.9 t^{2} - 36 t - 45 = 0$ .
Solve with the quadratic formula $t = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ , where $a = 4.9$ , $b = - 36$ , $c = - 45$ : $t = \frac{36 \pm 3 6 ^{2} + 4 ( 4.9 ) ( 45 )}{2 ( 4.9 )} \approx \frac{36 \pm 46.5}{9.8}$ . Discard the negative solution $t \approx - 1.1 s$ , so $t \approx 8.4 s$ .
Calculate horizontal distance: $x = v_{0 x} t = (77.0) (8.4) \approx 650 m$ .

Exam tip: If you get two positive solutions for t after solving the quadratic, check their context: one is when the projectile passes the target y-position on the way up, the other on the way down. Pick the solution that matches the problem's description of where the projectile lands.

5. Common Pitfalls (and how to avoid them)

Wrong move: Claiming the velocity of a projectile at maximum height is zero. Why: Students correctly note vertical velocity is zero at maximum height, but forget horizontal velocity remains constant for the entire flight. Correct move: When asked for total velocity at maximum height, remember it equals the constant initial horizontal velocity $v_{0 x}$ .
Wrong move: Using the level-ground range formula for a projectile launched from a cliff or into a valley. Why: Students memorize the simplified formula and forget its strict domain requirement that launch and landing elevation are equal. Correct move: Always confirm $y_{f ina l} = y_{ini t ia l}$ before using the simplified range formula; if not, solve the quadratic for t first.
Wrong move: Adding horizontal and vertical velocity as scalars ( $v = v_{x} + v_{y}$ ) to get total speed. Why: Students forget velocity is a vector, and add perpendicular components incorrectly. Correct move: Always calculate total speed as $v = v_{x}^{2} + v_{y}^{2}$ when combining perpendicular components.
Wrong move: Using radians instead of degrees on a calculator to calculate sine/cosine of launch angles. Why: Students forget to check calculator mode before starting calculations. Correct move: Confirm your calculator is set to degrees before starting any projectile problem, since launch angles are almost always given in degrees on the AP exam.
Wrong move: Adding acceleration to the horizontal velocity calculation. Why: Students overcomplicate problems and incorrectly assume gravity has a horizontal component. Correct move: Write $a_{x} = 0$ at the start of every projectile problem to remind yourself horizontal velocity is constant.
Wrong move: Mixing sign conventions for vertical acceleration (using $a_{y} = + g$ for upward positive). Why: Students encounter conflicting conventions in different resources and forget to adjust all terms consistently. Correct move: Write your chosen convention and the value of $a_{y}$ explicitly next to your coordinate system before starting calculations.

6. Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

A ball is thrown horizontally off the top of a 19.6 m tall building with an initial horizontal speed of 15 m/s. What is the time of flight of the ball before it hits the ground? Ignore air resistance. A) 1.0 s B) 2.0 s C) 4.0 s D) 15 s

Worked Solution: This is an uneven elevation projectile problem, so we solve for time from vertical motion. The ball is thrown horizontally, so initial vertical velocity $v_{0 y} = 0$ . Set origin at the throw point, so final $y = - 19.6 m$ . The y-equation becomes: $- 19.6 = 0 - \frac{1}{2} (9.8) t^{2}$ , which simplifies to $19.6 = 4.9 t^{2}$ , so $t^{2} = 4$ and $t = 2.0 s$ . Initial horizontal speed does not affect time of flight due to independence of motion. The correct answer is B.

Question 2 (Free Response)

A student throws a rock from the edge of a cliff of height $h$ with initial speed $v_{0}$ at an angle $θ$ above the horizontal. The rock lands on the ground below the cliff. (a) Derive an expression for the total time of flight $t$ of the rock in terms of $v_{0}$ , $θ$ , $g$ , and $h$ . (b) Explain why the time of flight does not depend on the horizontal component of initial velocity. (c) If the student wants to maximize the horizontal range of the rock, will they throw at an angle greater than, less than, or equal to 45°? Justify your answer.

Worked Solution: (a) Set origin at the launch point, positive y upward, so $y_{0} = 0$ , $y_{f ina l} = - h$ . Initial vertical velocity is $v_{0 y} = v_{0} sin θ$ . Substitute into the y-displacement kinematic equation: $- h = v_{0} sin θ \cdot t - \frac{1}{2} g t^{2}$ Rearrange into standard quadratic form: $\frac{1}{2} g t^{2} - v_{0} sin θ \cdot t - h = 0$ Solve with the quadratic formula and take the positive (physical) root: $t = \frac{v _{0} sin θ + ( v _{0} sin θ ) ^{2} + 2 g h}{g}$

(b) By the principle of independence of motion, gravity only accelerates the projectile in the vertical direction, so vertical motion (which determines how long the projectile is airborne) is completely unaffected by the horizontal component of velocity. The horizontal velocity only determines how far the projectile travels after it is airborne, not how long it is airborne.

(c) The maximum range occurs at an angle less than 45°. For level ground, maximum range is 45°, but here the projectile is launched above the landing elevation. Increasing the horizontal component of velocity (by decreasing the angle below 45°) gives a larger increase in range than increasing time of flight from a higher launch angle, so the optimal angle is less than 45°.

Question 3 (Application / Real-World Style)

An elite long jumper has a center of mass 1.0 m above the ground at takeoff, and launches off the board at 11.5 m/s at 18° above the horizontal. The jumper's center of mass is 0.25 m above the ground at landing. Estimate the total horizontal distance of the jump, measured from the takeoff board to landing.

Worked Solution: Set origin at the center of mass at takeoff, so $y_{f ina l} = 0.25 - 1.0 = - 0.75 m$ . Resolve initial velocity: $v_{0 y} = 11.5 sin 1 8^{\circ} \approx 3.56 m/s$ , $v_{0 x} = 11.5 cos 1 8^{\circ} \approx 10.9 m/s$ . Substitute into the y-equation: $- 0.75 = 3.56 t - 4.9 t^{2}$ , which rearranges to $4.9 t^{2} - 3.56 t - 0.75 = 0$ . Solve the quadratic: $t = \frac{3.56 + 3.5 6 ^{2} + 4 ( 4.9 ) ( 0.75 )}{2 ( 4.9 )} \approx 0.91 s$ . Total horizontal distance: $x = v_{0 x} t = (10.9) (0.91) \approx 9.9 m$ . In context: This result matches the real-world performance of elite men's long jumpers, who regularly achieve jump distances between 8 and 10 meters.

7. Quick Reference Cheatsheet

Category	Formula	Notes
Core Principle	Horizontal and vertical motions are independent	Applies to all projectiles with no air resistance
Initial Velocity Components	$v_{0 x} = v_{0} cos θ_{0}$ , $v_{0 y} = v_{0} sin θ_{0}$	$θ_{0}$ = launch angle above horizontal
Acceleration Components	$a_{x} = 0$ , $a_{y} = - g = - 9.8 m/s^{2}$	Positive y = upward convention, no air resistance
Horizontal Kinematic Equation	$x = x_{0} + v_{0 x} t$	Always true for projectiles, since $a_{x} = 0$
Vertical Kinematic Equation	$y = y_{0} + v_{0 y} t - \frac{1}{2} g t^{2}$	Always true for projectiles near Earth's surface
Level Ground Time of Flight	$t = \frac{2 v _{0} s i n θ _{0}}{g}$	Only valid when $y_{f ina l} = y_{ini t ia l}$
Level Ground Maximum Height	$H = \frac{v _{0}^{2} s i n ^{2} θ _{0}}{2 g}$	Maximum height above launch elevation, level ground only
Level Ground Range	$R = \frac{v _{0}^{2} s i n 2 θ _{0}}{g}$	Maximum range at $θ_{0} = 4 5^{\circ}$ , level ground only
Total Speed at Time $t$	$v = v_{0 x}^{2} + (v_{0 y} - g t)^{2}$	Always true, combines perpendicular velocity components

8. What's Next

Projectile motion is the first core 2D kinematics topic you encounter in AP Physics 1, and it is a foundational prerequisite for all future topics involving motion in multiple dimensions. Next, you will extend the principles of vector resolution and independent motion to analyze forces in 2D, including inclined planes and static equilibrium problems. Later, projectile motion principles underpin analysis of uniform circular motion, orbital motion, and collision problems where momentum is conserved across two dimensions. Without mastering the core principle of independent perpendicular motions and consistent sign conventions for 2D kinematics, you will struggle to correctly set up equations for all these higher-level topics. Follow the links below for related topics in the AP Physics 1 syllabus:

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Projectile Motion — AP Physics 1 Study Guide

1. What Is Projectile Motion?

2. Independence of Perpendicular Motions

Worked Example

3. Projectiles Launched From Level Ground

Worked Example

4. Projectiles Launched From Uneven Elevation

Worked Example

5. Common Pitfalls (and how to avoid them)

6. Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

Question 2 (Free Response)

Question 3 (Application / Real-World Style)

7. Quick Reference Cheatsheet

8. What's Next

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