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

Concept: Angular momentum about a point: \[ L = m v_{\perp} \cdot r \] At highest point, velocity is horizontal.

Step 1: Velocity at highest point.
\[ v_x = v\cos45^\circ = \frac{v}{\sqrt{2}} \]

Step 2: Coordinates of highest point.
Time to reach highest point: \[ t = \frac{v\sin45^\circ}{g} = \frac{v}{\sqrt{2}g} \] Horizontal distance: \[ x = v_x \cdot t = \frac{v}{\sqrt{2}} \cdot \frac{v}{\sqrt{2}g} = \frac{v^2}{2g} \] Height: \[ y = \frac{v^2 \sin^2 45^\circ}{2g} = \frac{v^2}{4g} \]

Step 3: Angular momentum.
Perpendicular distance from origin to velocity line = \(y\) \[ L = m v_x \cdot y = m \cdot \frac{v}{\sqrt{2}} \cdot \frac{v^2}{4g} = \frac{mv^3}{4\sqrt{2}g} \]