Lecture Notes on Projectile Motion

Introduction to Projectile Motion

• Main Feature: Projectile launched horizontally
• Key Points:
  • Initial Vertical Velocity = 0
  • Horizontal velocity remains constant throughout motion.

Key Equations and Concepts

• Common Quantity: Time is the same for both horizontal and vertical motions.
• Displacement Equation:
  • ( s = ut + \frac{1}{2}at^2 )
  • For horizontal launch:
    • ( s = 0 + \frac{1}{2} * 9.81 * t^2 )

Calculation Examples

Example: Cannonball Launch

• Given:
  • Horizontal velocity = 12 m/s
  • Vertical acceleration = -9.81 m/s²
• Time Calculation:
  • Using ( t = 1.43 ) seconds
• Distance Calculation:
  • Horizontal distance = speed × time = 12 m/s × 1.43 s = 17.16 m

Velocity Components

• Vertical Component:
  • Use vector triangle for resultant velocity.
  • Horizontal component = 12 m/s
  • Vertical component = 14 m/s
• Resultant Velocity:
  • ( v = \sqrt{12^2 + 14^2} \approx 18.4 ) m/s
  • Direction Calculation:
    • Use ( \tan(\theta) = \frac{opposite}{adjacent} )

Types of Motion

• Single vs. Double Motion:
  • Projectile can be a combination of horizontal and vertical motions.
  • Total vertical displacement is negative (below launch point).

Example: Launch from Height

• Given: Launch at 15 m/s at 30 degrees
• Vertical and Horizontal Components:
  • Vertical = 15 sin(30)
  • Horizontal = 15 cos(30)
• Time Calculation:
  • Solve using quadratic formula for vertical motion.
• Final Calculations:
  • Distance traveled = speed × time = 15 cos(30) × time
  • Result: Distance = 29.2 m

Conclusion

• Remember: Both horizontal and vertical motions are interconnected through time.
• Practice: Work through additional examples for mastery.