Question 1• 5 Marks
Solution:
Given Data:
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Target / To Prove:
Formula Applied:
Step-by-Step Working:
Note: Since t = 2.0 s < t_up (3.53 s), the projectile is still ascending at this instant.
Unit 3 • Class 11Physics (Grade 11) (Phy.101)
Kinematics & Projectile Motion
शुद्ध गतिशास्त्र र प्रक्षेप्य गति
Equations of motion under uniform acceleration, relative velocity, displacement-time and velocity-time graphs, projectile motion equations.
Lesson Summary & Key Concepts
Rigorous derivation of oblique projectile trajectory, parabolic path equation, time of flight, maximum height, horizontal range, and condition for maximum range.
Textbook Exercises & Solutions
Showing 1 of 1 Solved QuestionsQuick Revision Takeaways
Common Exam Pitfalls to Avoid
⚠️ Common Error: Assuming the velocity of a projectile at its peak altitude is zero.
✓ Correct Approach: Only the vertical component v_y = 0; horizontal component v_x = u cos θ remains constant.
Why: Kinetic energy at maximum height is (1/2)m(u cos θ)², not zero.
Content Provenance & Verification
Editorial Verified & Published
Original Author / ContributorNEB Physics Subject Specialist
Verified BySenior Subject Specialist (TU / NEB Panel)
Curriculum Versionv1.0 (CDC 2078/2080 Grid)
Last Verified Date2026-10-23
Legitimate Official Curriculum References
public curriculum
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