MP Board · Class 11 · Physics · Motion in a PlaneThe maximum range of a projectile is $100\text{ m}$. What is the maximum height reached by it for this launch condition?
Step-by-Step Solution
Maximum range occurs at $\theta = 45^\circ$ and is given by $R_{\max} = \frac{u^2}{g} = 100\text{ m}$. The maximum height at this angle is $H = \frac{u^2 \sin^2(45^\circ)}{2g} = \frac{u^2}{4g} = \frac{100}{4} = 25\text{ m}$.
Detailed Options Breakdown
Option : $25\text{ m}$ (Correct Answer)
Correct choice. Refer to the step-by-step verified solution guidelines above for details.
Option 1: $50\text{ m}$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Motion in a Plane.
Option 2: $75\text{ m}$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Motion in a Plane.
Option 3: $100\text{ m}$
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.