Question

A particle of mass m is projected with a velocity v at an angle of \(45^\circ\) with horizontal. When the particle is at its maximum height, the magnitude of its angular momentum about the point of projection is:

• A. zero
• B. \( \frac{mv^3}{4\sqrt{2}g} \)
• C. \( \frac{mv^3}{\sqrt{2}g} \)
• D. \( \frac{mv^3}{\sqrt{2}g} \)

Hint:

At maximum height, velocity is horizontal. Use perpendicular distance (height) for angular momentum.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Angular momentum \( \vec{L} \) is defined as \( \vec{r} \times \vec{p} \).
At the maximum height, the particle moves purely horizontally, so its momentum is horizontal. The perpendicular distance to the point of projection is the maximum height \( H \).
: Key Formula or Approach:
1. \( L = p \cdot r_{\perp} \).
2. Velocity at max height \( v_x = v \cos \theta \).
3. Max height \( H = \frac{v^2 \sin^2 \theta}{2g} \).
Step 2: Detailed Explanation:
Given \( \theta = 45^\circ \).
- Momentum at max height:
\[ p = m v_x = m v \cos 45^\circ = \frac{mv}{\sqrt{2}} \]
- Perpendicular distance from origin (point of projection) to the line of motion at max height is the vertical height \( H \):
\[ H = \frac{v^2 \sin^2 45^\circ}{2g} = \frac{v^2 (1/2)}{2g} = \frac{v^2}{4g} \]
- Magnitude of Angular Momentum \( L \):
\[ L = p \times H \]
\[ L = \left( \frac{mv}{\sqrt{2}} \right) \times \left( \frac{v^2}{4g} \right) \]
\[ L = \frac{mv^3}{4\sqrt{2}g} \]
Step 3: Final Answer:
The magnitude of its angular momentum is \( \frac{mv^3}{4\sqrt{2}g} \).