Lectures on Projectile Motion on an Inclined Plane

Introduction

• Today's lecture is focused on JEE syllabus as NEET syllabus does not cover projection on an inclined plane.
• NEET syllabus for this chapter was completed in the last lecture.
• Topics for today:
  • Motion of an object projectile on an inclined plane at an angle θ.
  • Range, time of flight, and maximum height on an inclined plane.
  • Two cases: from bottom to top and from top to bottom.
  • Numerical problems related to the concept.

Case 1: From Bottom to Top

• Setup: Inclined plane with angle α, initial velocity u at angle θ.
  • Angle θ is the angle of projection from the inclined plane.
  • Angle of projection with the ground is α + θ.
• Velocity Components:​
  • Initial velocity along x-direction: u cos(θ + α).
  • Initial velocity along y-direction: u sin(θ + α).
• Simplifying approach: Instead of traditional x, y axes, use axes along and perpendicular to the inclined plane:
  • New x-axis along the inclined plane.
  • New y-axis perpendicular to the inclined plane.
• Modified Velocity Components:​
  • Along inclined plane: u cos(θ).
  • Perpendicular to inclined plane: u sin(θ).
• Acceleration due to gravity components:
  • Along inclined plane: g sin(α).
  • Perpendicular to inclined plane: g cos(α).
• Equations of Motion:​
  • x-direction: u cos(θ) - g sin(α)t.
  • y-direction: u sin(θ) - g cos(α)t.
• Time of Flight (T):
  • Derived using second equation of motion perpendicular to inclined plane:
  • T = (2u sin(θ)) / (g cos(α))
• Maximum Height (H):
  • Derived using third equation of motion perpendicular to inclined plane:
  • H = (u² sin²(θ)) / (2g cos(α))
• Range (R):​
  • Derived using second equation of motion along inclined plane:
  • R = (u² sin(2θ)) / (g cos²(α))
• Angle for Maximum Range:
  • Angle θ for max range: (90° - α)/ 2
  • Maximum range: u² / (g (1 + sin(α)))

Case 2: From Top to Bottom

• Setup: Similar to case 1, but projecting from top to bottom.
• Initial and final velocities:
  • Initial velocity components:
    • Along x-direction: u cos(θ).
    • Along y-direction: u sin(θ).
• Key Differences in Components:
  • Velocity and acceleration components in opposite directions compared to case 1.
• Time of Flight (T):​
  • Same as case 1: T = (2u sin(θ)) / (g cos(α))
• Maximum Height (H):
  • Same as case 1: H = (u² sin²(θ)) / (2g cos(α))
• Range (R):
  • Similar derivation with positive acceleration component:
  • R = (u² sin(2θ)) / (g cos²(α))
• Angle for Maximum Range:
  • θ for max range: (45° + α)/ 2
  • Maximum range using derived formula.

Numericals

Problem 1:

• Bullet projection:​ Fired with 50 m/s at 37° from an inclined plane of 30°.
• Calculations:
  1. Initial Velocities:​
  • u x = 40 m/s, u y = 30 m/s
  2. Maximum Height:
  • H = 30√3 meters
  3. Time of Flight:​
  • T = 4√3 seconds
  4. Range:​
  • R = 40(√3 - 1) meters
  5. Angle for Maximum Range:
  • θ = 30°
  6. Maximum Range:
  • = 500/3 meters

Problem 2 (Homework):

• Projectile from top to bottom:
  • Derive all required parameters.

Problem 3:

• Projectile perpendicular to plane:
  • Initial velocity v and plane at 30°.
  • Required: Time of flight and range.
  • Time of flight: 2v / g cos(α)
  • Range: v² / (g sin(2α))

Problem 4:

• Two inclined planes:
  • 30° and 60° angles.
  • Projectile with 10√3 m/s at right angle.
  • Required: Time of flight, impact speed, distances.
  • Time of flight: 2 seconds.
  • Speed at impact: 10 m/s
  • Distance PQ = 20 m
  • Height calculations using trigonometry.

Summary

• Detailed discussion on projections on inclined planes.
• Important derivations for JEE syllabus.
• Covered two cases of projection.
• Extensive numerical problem-solving.