NCERT · Class 11 · Physics · Motion in a PlaneAt what angle of projection is the horizontal range of a projectile maximum for a given initial velocity?
Step-by-Step Solution
The formula for horizontal range is $R = \frac{u^2 \sin(2\theta)}{g}$. The range is maximum when $\sin(2\theta) = 1$, which implies $2\theta = 90^\circ$, so $\theta = 45^\circ$.
Detailed Options Breakdown
Option : $30^\circ$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Motion in a Plane.
Option 1: $45^\circ$ (Correct Answer)
Correct choice. Refer to the step-by-step verified solution guidelines above for details.
Option 2: $60^\circ$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Motion in a Plane.
Option 3: $90^\circ$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Motion in a Plane.
💡 Study Guide: This question tests core syllabus concepts from Motion in a Plane. For formulas, key summaries, and mock exam reference guides, read the full Motion in a Plane Revision Notes.