Projectile Motion

Q: A ball is launched at 20 m/s at an angle of 30°. What is its horizontal velocity (v_x) after 2 seconds?

17.3 m/s v_x = v_0 \cos(\theta) = 20 \cos(30°) = 20 (\sqrt{3}/2) \approx 17.3 m/s. Since horizontal acceleration is zero, v_x remains constant throughout the flight.

Q: At what point in a projectile's trajectory is the vertical velocity (v_y) zero?

At the peak (maximum height). The vertical component of velocity decreases as the object rises, reaches zero at the highest point, and then increases in the downward direction as it falls. This is the moment when the projectile transitions from upward to downward motion.

Q: A ball is launched at 30° and 60° with the same initial velocity. Which launch angle results in a greater horizontal range?

They are the same. Complementary angles (30° and 60° add to 90°) produce the same horizontal range because \sin(2 \times 30°) = \sin(60°) and \sin(2 \times 60°) = \sin(120°) = \sin(60°). However, 60° will result in a higher peak and longer time in the air.

Q: A projectile is launched horizontally from a cliff 80 m high with initial speed 20 m/s. How far from the base of the cliff does it land? (Use g = 10 m/s²)

80 m First find time of flight using vertical motion: y = \frac{1}{2}gt^2, so 80 = \frac{1}{2}(10)t^2, giving t = 4 s. Then horizontal distance: x = v_x t = 20(4) = 80 m.

Visualize 2D projectile trajectories with adjustable launch angle, initial velocity, and gravitational acceleration. Observe how each parameter affects range, maximum height, and time of flight in real time.

GUIDED INQUIRY

Analyze the simulation using this focus prompt: "Why does the horizontal velocity vector stay the same length while the vertical one changes?"

WHAT IS PROJECTILE MOTION?

Projectile motion is the motion of an object thrown or projected into the air, subject only to the acceleration of gravity. The key insight for AP Physics 1 is the **independence of horizontal and vertical motion**: gravity only affects the vertical component (acceleration m/s²), while the horizontal velocity remains constant (neglecting air resistance). This results in a parabolic trajectory that can be analyzed using two separate sets of kinematic equations. For any projectile launched at angle with initial speed : - Horizontal component: (remains constant) - Vertical component: (changes due to gravity)

KEY FORMULAS AND DERIVATIONS

**Maximum Height**: The projectile reaches maximum height when . Using , we get: h_{max} = \frac{v_{y0}^2}{2g} = \frac{(v_0\sin\theta)^2}{2g} **Time of Flight**: For a projectile landing at the same height it was launched, the total time is: t_{total} = \frac{2v_{y0}}{g} = \frac{2v_0\sin\theta}{g} **Range**: The horizontal distance traveled is: R = v_{x0} \cdot t_{total} = \frac{v_0^2\sin(2\theta)}{g} Note: Maximum range occurs at because is the maximum value.

HOW TO USE THIS VISUALIZATION

1. **Adjust the Launch Angle**: Use the slider to change the initial direction. Notice how maximizes the horizontal range for level ground. 2. **Change Initial Velocity**: Observe how doubling the velocity quadruples both the maximum height and range (both depend on ). 3. **Toggle Vectors**: Enable the velocity and acceleration vectors to see how changes while stays constant. 4. **Compare Complementary Angles**: Launch at and with the same speed—they produce the same range but different trajectories.

CORE FORMULAS

Horizontal displacement (constant velocity)

Vertical velocity (constant acceleration)

Vertical displacement

Range formula (level ground)

Maximum height

AP EXAM CONNECTION

Unit: Unit 1: Kinematics (Topic 1.3)
Learning Objective: 1.3.1

COMMON MISCONCEPTIONS

Thinking horizontal and vertical motions affect each other.
Believing the horizontal velocity changes during flight (it remains constant without air resistance).
Assuming velocity is zero at the peak (only the vertical component is zero; horizontal velocity is still present).
Confusing the angle that maximizes range (45°) with the angle that maximizes height (90°).

KEY TAKEAWAYS

Horizontal and vertical motions are independent and can be analyzed separately.
Horizontal velocity remains constant (no horizontal acceleration).
Vertical acceleration is constant ( m/s² downward).
The trajectory is a parabola.
Maximum range on level ground is achieved at 45° launch angle.
Complementary angles (e.g., 30° and 60°) produce the same range but different flight times and maximum heights.

PRACTICE QUESTIONS

Q1 (QUANTITATIVE): A ball is launched at 20 m/s at an angle of 30°. What is its horizontal velocity () after 2 seconds?

Show Answer & Explanation

Answer: 17.3 m/s

Explanation: m/s. Since horizontal acceleration is zero, remains constant throughout the flight.

Q2 (CONCEPTUAL): At what point in a projectile's trajectory is the vertical velocity () zero?

Show Answer & Explanation

Answer: At the peak (maximum height).

Explanation: The vertical component of velocity decreases as the object rises, reaches zero at the highest point, and then increases in the downward direction as it falls. This is the moment when the projectile transitions from upward to downward motion.

Q3 (CONCEPTUAL): A ball is launched at and with the same initial velocity. Which launch angle results in a greater horizontal range?

Show Answer & Explanation

Answer: They are the same.

Explanation: Complementary angles ( and add to ) produce the same horizontal range because and . However, will result in a higher peak and longer time in the air.

Q4 (QUANTITATIVE): A projectile is launched horizontally from a cliff 80 m high with initial speed 20 m/s. How far from the base of the cliff does it land? (Use m/s²)

Show Answer & Explanation

Answer: 80 m

Explanation: First find time of flight using vertical motion: , so , giving s. Then horizontal distance: m.