Projectile Motion

Introduction

Projectile motion is 2D motion under the influence of gravity only (ignoring air resistance).

Examples:

• Ball thrown at an angle
• Cannonball fired from a cannon
• Water from a fountain
• Long jumper

Key Characteristics

1. Horizontal motion: constant velocity (no horizontal acceleration) $a_x = 0$ $v_x = v_{0x} = \text{constant}$

2. Vertical motion: constant acceleration due to gravity $a_y = -g = -9.8 \text{ m/s}^2$

3. Parabolic path: The trajectory is a parabola

4. Independence: Horizontal and vertical motions are independent

Initial Velocity Components

If launched at angle $\theta_0$ with initial speed $v_0$:

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

Equations of Motion

Horizontal Direction (constant velocity)

$x = x_0 + v_{0x}t$ $v_x = v_{0x} = \text{constant}$

Vertical Direction (constant acceleration)

$y = y_0 + v_{0y}t - \frac{1}{2}gt^2$ $v_y = v_{0y} - gt$ $v_y^2 = v_{0y}^2 - 2g(y - y_0)$

Note: Using $-g$ for downward acceleration.

Key Features of Parabolic Trajectory

At the Peak

• Vertical velocity: $v_y = 0$ (changes from positive to negative)
• Horizontal velocity: $v_x = v_{0x}$ (unchanged)
• Total velocity: $v = v_x$ (purely horizontal)
• Acceleration: $a = -g$ (always downward, even at peak!)

Symmetry Properties

For projectile launched and landing at same height:

1. Time up = Time down $t_{up} = t_{down} = \frac{v_{0y}}{g}$

2. Total time of flight: $t_{total} = \frac{2v_{0y}}{g} = \frac{2v_0 \sin \theta_0}{g}$

3. Maximum height: $h_{max} = \frac{v_{0y}^2}{2g} = \frac{(v_0 \sin \theta_0)^2}{2g}$

4. Range (horizontal distance): $R = \frac{v_0^2 \sin(2\theta_0)}{g}$

5. Landing speed = Launch speed (magnitude)

6. Landing angle = Launch angle (below horizontal)

Special Cases

Horizontal Launch ($\theta_0 = 0°$)

• $v_{0x} = v_0$
• $v_{0y} = 0$
• Time in air: $t = \sqrt{\frac{2h}{g}}$ where $h$ is initial height

Vertical Launch ($\theta_0 = 90°$)

• $v_{0x} = 0$ (no horizontal motion)
• $v_{0y} = v_0$
• 1D motion only

45° Launch (Maximum Range)

• For a given speed $v_0$, $\theta_0 = 45°$ gives maximum range
• $R_{max} = \frac{v_0^2}{g}$

Complementary Angles

• $\theta_0$ and $(90° - \theta_0)$ give the same range
• Example: 30° and 60° both give same range (but different trajectories!)

Problem-Solving Strategy

1. Draw a diagram showing trajectory
2. Set up coordinate system (origin at launch point)
3. List knowns: $v_0$, $\theta_0$, $x_0$, $y_0$, etc.
4. Break $v_0$ into components: $v_{0x}$ and $v_{0y}$
5. Write separate equations for $x$ and $y$
6. Use time as the link between horizontal and vertical
7. Solve step by step

Common Questions

Q: What is the acceleration at the peak? A: $a = -g$ downward. Acceleration is always $-g$ throughout flight!

Q: What is the velocity at the peak? A: $v = v_x = v_{0x}$ (horizontal only, since $v_y = 0$)

Q: How do you find time to reach maximum height? A: Use $v_y = v_{0y} - gt$ and set $v_y = 0$: $t = \frac{v_{0y}}{g}$

Q: How do you find range? A: Find total time of flight, then $R = v_x \cdot t_{total}$

Real-World Applications

• Sports: basketball, soccer, golf, baseball
• Military: ballistics, artillery
• Engineering: water fountains, sprinkler systems
• Space: satellite trajectories (approximately)