Lecture Notes on Projectile Motion

Ball Rolling Off a Cliff

Scenario

• A ball rolls off a 300-meter cliff with a horizontal speed of 20 m/s and hits the ground.
• Points of interest: Point A (top of the cliff) and Point B (ground level).

Part A: Time of Flight

• Formula: [ h = \frac{1}{2} g t^2 ]
• Rearranged: [ t = \sqrt{\frac{2h}{g}} ]
• Values:
  • Height (h) = 300 m
  • Gravity (g) = 9.8 m/s²
• Calculation:
  • [ t = \sqrt{\frac{2 \times 300}{9.8}} \approx 7.82 ext{ seconds} ]

Part B: Final Velocity Before Hitting the Ground

• Concept: Need to find the final velocity vector (magnitude and direction).
• Magnitude Formula: [ v_f = \sqrt{v_{i}^2 + g t^2} ]
• Values:
  • Initial horizontal speed (v_i) = 20 m/s
  • Time (t) = 7.82 s
• Calculation:
  • [ v_f = \sqrt{20^2 + (9.8 \times 7.82)^2} \approx 79.25 ext{ m/s} ]

Finding the Angle

• Components:
  • Horizontal Component (Vx) = 20 m/s
  • Vertical Component (Vy) = -76.68 m/s (calculated using -gt)
• Angle Calculation:
  • [ \theta = \arccos(\frac{V_x}{V_f}) ]
  • [ \theta = \arccos(\frac{20}{79.25}) \approx 75.38° ] (below the x-axis)

Ball Kicked from the Ground

Scenario

• A ball is kicked from the ground at an angle of 30° with an initial speed of 50 m/s.

Part A: Time of Flight

• Formula: [ \text{Time of flight} = \frac{2V \sin(\theta)}{g} ]
• Calculation:
  • [ \text{Time} = \frac{2 \times 50 \times \sin(30)}{9.8} \approx 5.10 ext{ seconds} ]

Part B: Final Velocity

• Final Velocity at Point C:
  • Magnitude: 50 m/s (same as initial speed, just direction changes)

Verification

• Using the Final Velocity formula:
  • [ v_f = \sqrt{v_i^2 - 2gtV_i \sin(\theta) + gt^2} ]
• Result: Approximately 50 m/s, confirming the magnitude.

Ball Kicked from a 400-Meter Cliff

Scenario

• A ball is kicked from a 400-meter cliff at an angle of 30° with an initial speed of 80 m/s.

Part A: Time of Flight

• Formula: [ t = \frac{V_i \sin(\theta) + \sqrt{V_i^2 \sin^2(\theta) + 2gy}}{g} ]
• Calculation:
  • Plugging in values gives approximately 14 seconds.

Part B: Final Velocity Before Hitting the Ground

• Magnitude Calculation:
  • [ v_f = \sqrt{80^2 - 2 \times 9.8 \times 14 \times 80 \times \sin(30) + 9.8 \times 14^2} \approx 119.36 ext{ m/s} ]

Angle Calculation

• Angle Below the Horizontal:
  • [ \theta = \arccos(\frac{V_x}{V_f}) ]
  • Resulting angle is approximately 54.5° below the horizontal.

Summary

• Importance of using formulas in projectile motion to quickly solve problems.
• Formula sheet provided as a helpful reference for calculations.