AP Physics 1 Projectile Motion

AP Physics 1· AP Physics 1 CED — Kinematics· 14 min read

1. Core Principles of Projectile Motion★★☆☆☆⏱ 4 min

Projectile motion describes the motion of an object (called a projectile) launched into the air, where after launch, the only force acting on it is constant downward gravity (air resistance is always ignored in AP Physics 1 problems). It is assessed in both multiple-choice and free-response sections of the AP exam, often as part of multi-concept problems linking to forces, energy, or momentum.

📘 Definition

Projectile Motion

$v_{0}$ = initial speed, $θ_{0}$ = launch angle

Motion of an object under the sole influence of constant gravitational acceleration, with air resistance neglected.

Example:

A kicked soccer ball, a thrown rock, a cannonball

The foundational principle of projectile motion is that horizontal and vertical motions are completely independent, connected only by the shared time of flight. Gravity acts exclusively vertically, so there is no horizontal acceleration, allowing us to split any 2D problem into two separate 1D kinematics problems.

v_{0x} = v_0 \cos\theta_0 \qquad v_{0y} = v_0 \sin\theta_0

With the standard convention (origin at launch, positive y upward, positive x in direction of motion): acceleration components are $a_{x} = 0$ , $a_{y} = - g$ , where $g = 9.8 m/s^{2}$ . The kinematic equations simplify to:

Horizontal motion: $x = x_{0} + v_{0 x} t$ (constant velocity)
Vertical motion: $v_{y} = v_{0 y} - g t$ , $y = y_{0} + v_{0 y} t - \frac{1}{2} g t^{2}$ , $v_{y}^{2} = v_{0 y}^{2} - 2 g (y - y_{0})$

📐 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?

1
Set origin at launch, so $x_{0} = 0$ , $y_{0} = 0$ , $t = 1.0 s$ .
2
Resolve initial velocity into components:
3
$v_{0x} = 20\cos30^\circ = 10\sqrt{3} \approx 17.3\ \text{m/s}, \quad v_{0y} = 20\sin30^\circ = 10\ \text{m/s}$
4
Calculate horizontal position:
5
$x = v_{0x}t = (17.3)(1.0) = 17.3 \approx 17\ \text{m}$
6
Calculate vertical position:
7
$y = v_{0y}t - \frac{1}{2}gt^2 = (10)(1) - (4.9)(1^2) = 5.1\ \text{m}$
8
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, marking the positive y direction, to eliminate 90% of common sign errors on the AP exam.

2. Projectiles Launched From Level Ground★★☆☆☆⏱ 3 min

When a projectile launches and lands at the same vertical elevation ( $y_{f ina l} = y_{ini t ia l}$ ), we can derive simplified expressions for time of flight, maximum height, and range that speed up multiple-choice calculations.

🔬 Derivation

Goal:

Derive simplified formulas for symmetric level-ground projectile motion

Starting from:

Vertical displacement equation $y - y_{0} = 0$

1
Set $y - y_{0} = 0$ in the vertical displacement equation:
2
$0 = v_{0y}t - \frac{1}{2}gt^2 = t\left(v_{0y} - \frac{1}{2}gt\right)$
3
The non-trivial solution (excluding launch at $t = 0$ ) gives total time of flight:
4
$t_{total} = \frac{2v_{0y}}{g} = \frac{2v_0\sin\theta_0}{g}$
5
Maximum height occurs when vertical velocity $v_{y} = 0$ , at half the total time. Substituting back gives maximum height:
6
$H = \frac{v_0^2\sin^2\theta_0}{2g}$
7
Range (total horizontal distance) is $R = v_{0 x} t_{t o t a l}$ , which simplifies to:
8
$R = \frac{v_0^2 \sin2\theta_0}{g}$

Result:

Maximum range for level-ground launch occurs at $θ_{0} = 4 5^{\circ}$ , since $sin 2 θ$ reaches its maximum value of 1 at $2 θ = 9 0^{\circ}$ .

📐 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?

1
Confirm launch and landing are at the same elevation, so simplified formulas apply.
2
Calculate maximum height:
3
$H = \frac{v_0^2\sin^2\theta_0}{2g} = \frac{(22^2)(\sin35^\circ)^2}{2(9.8)} \approx 8.1\ \text{m}$
4
Calculate total range:
5
$R = \frac{v_0^2\sin2\theta_0}{g} = \frac{484(\sin70^\circ)}{9.8} \approx 46\ \text{m}$
6
Final result: Maximum height ≈ 8.1 m, total range ≈ 46 m.

Exam tip:

The simplified range formula only works for level ground (equal launch and landing elevation). Never use it for projectiles launched from cliffs or hills unless you confirm elevations are equal.

3. Projectiles Launched From Uneven Elevation★★★☆☆⏱ 4 min

Most non-routine AP Physics 1 projectile problems involve launch and landing at different elevations, such as throwing a ball off a building or launching a projectile at a target on a hill. Simplified level-ground formulas do not apply here, so we always return to the core kinematic equations and solve for time first.

Write the vertical displacement equation using the known final y position
Rearrange into standard quadratic form in $t$
Discard any non-physical negative time solution
Use the valid positive time to solve for unknown horizontal distance or velocity

📐 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 flat ground below the cliff. What is the horizontal distance from the base of the cliff to where the cannonball lands?

1
Set origin at the cannon, positive y upward, so $y_{0} = 0$ , final $y = - 45 m$ . Resolve initial velocity components:
2
$v_{0x} = 85\cos25^\circ \approx 77.0\ \text{m/s}, \quad v_{0y} = 85\sin25^\circ \approx 36.0\ \text{m/s}$
3
Substitute into the vertical displacement equation and rearrange to quadratic form:
4
$-45 = 36t - 4.9t^2 \implies 4.9t^2 - 36t - 45 = 0$
5
Solve with the quadratic formula $t = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ , where $a = 4.9$ , $b = - 36$ , $c = - 45$ :
6
$t = \frac{36 \pm \sqrt{36^2 + 4(4.9)(45)}}{9.8} \approx \frac{36 \pm 46.5}{9.8}$
7
Discard the negative solution $t \approx - 1.1 s$ , so valid time is $t \approx 8.4 s$ .
8
Calculate total horizontal distance:
9
$x = v_{0x}t = (77.0)(8.4) \approx 650\ \text{m}$

Exam tip:

If you get two positive solutions for $t$ , one corresponds to passing the target y-position on the way up, and the other on the way down. Pick the solution that matches the problem's description of where the projectile lands.

4. AP-Style Concept Check★★★☆☆⏱ 3 min

✓ Quick check

Test your understanding of core projectile motion concepts with these AP-style questions:

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.
- 1.0 s
- 2.0 s
- 4.0 s
- 15 s
Reveal answer
1 —
Time of flight depends only on vertical motion. With $v_{0 y} = 0$ , $y = - 19.6 = - 4.9 t^{2} ⟹ t^{2} = 4 ⟹ t = 2.0 s$ . Horizontal speed does not affect time of flight due to independence of motion.
A rock is thrown from the top of a cliff at an angle above the horizontal, landing on the ground below. To maximize the horizontal range, what launch angle should you use?
- Equal to 45°
- Greater than 45°
- Less than 45°
- Range is the same for all angles
Reveal answer
2 —
For uneven elevation where launch is higher than landing, maximum range occurs at an angle less than 45°, because increasing the horizontal component of velocity gives a larger gain in range than increasing time of flight from a higher launch angle.

5. Common Pitfalls

Wrong move:

Claiming total velocity of a projectile at maximum height is zero

Why:

Students correctly remember vertical velocity is zero at maximum height, but forget horizontal velocity remains constant throughout the entire flight

Correct move:

At maximum height, total velocity equals the constant initial horizontal velocity $v_{0 x}$

Wrong move:

Using the level-ground range formula for projectiles launched from uneven elevation

Why:

Students memorize the simplified formula and forget it only applies when 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 time first

Wrong move:

Adding horizontal and vertical velocity as scalars to get total speed ( $v = v_{x} + v_{y}$ )

Why:

Students forget velocity is a vector, and add perpendicular components incorrectly

Correct move:

Calculate total speed using the Pythagorean theorem: $v = v_{x}^{2} + v_{y}^{2}$

Wrong move:

Using radians instead of degrees on a calculator for sine/cosine of launch angles

Why:

Students forget to check calculator mode before starting calculations, leading to incorrect numerical results

Correct move:

Confirm your calculator is set to degrees before starting any projectile problem; launch angles are almost always given in degrees on the AP exam

Wrong move:

Adding non-zero 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, using $a_{y} = + g$ when positive y is upward

Why:

Students encounter conflicting conventions from different sources 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. Quick Reference Cheatsheet

Category	Formula	Notes
Core Principle	Horizontal and vertical motions are independent	All projectiles, 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 Kinematics	$x = x_{0} + v_{0 x} t$	Always true for projectiles
Vertical Kinematics	$y = y_{0} + v_{0 y} t - \frac{1}{2} g t^{2}$	Always true 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, level ground only
Level Ground Range	$R = \frac{v _{0}^{2} s i n 2 θ _{0}}{g}$	Max range at 45°, level ground only
Total Speed at Time $t$	$v = v_{0 x}^{2} + (v_{0 y} - g t)^{2}$	Always true, combines perpendicular components

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Projectile maximum height calculation
2022 · FRQ
Uneven elevation projectile range

What's Next

Projectile motion is the first core 2D kinematics topic you learn in AP Physics 1, and it is a foundational prerequisite for all future topics involving motion in multiple dimensions. The core principle of independent perpendicular motion and consistent sign conventions you practice here will be used repeatedly in later units. Next, you will extend these principles of vector resolution to analyze forces in 2D, including inclined planes and static equilibrium. Later, projectile motion concepts underpin analysis of uniform circular motion, orbital motion, and 2D collision problems where momentum is conserved. Mastering this topic now will make all higher-dimensional motion problems much easier to set up and solve correctly on the exam.