Question

A particle of mass \( m \) is projected with a velocity \( u \) making an angle of \( 30^\circ \) with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height \( h \) is:

• A. \( \frac{\sqrt{3}}{16} \frac{m u^3}{g} \)
• B. \( \frac{\sqrt{3}}{2} \frac{m u^2}{g} \)
• C. \( \frac{m u^3}{\sqrt{2} g} \)
• D. zero

Hint:

For projectile motion, the angular momentum at the highest point is calculated as \( L = m v_x H \), where \( v_x \) is the horizontal velocity and \( H \) is the maximum height.

Answer 1

The Correct Option is A

Step 1: {Define angular momentum} 
Angular momentum at maximum height is given by: \[ L = m v H \] where \( H \) is the maximum height. 
Step 2: {Find the horizontal velocity} 
The horizontal component of velocity remains constant: \[ v = u \cos 30^\circ = u \times \frac{\sqrt{3}}{2} \] 
Step 3: {Find the maximum height} 
Using the kinematic equation: \[ H = \frac{u^2 \sin^2 30^\circ}{2g} \] Substituting \( \sin 30^\circ = \frac{1}{2} \): \[ H = \frac{u^2 \times \left(\frac{1}{2}\right)^2}{2g} = \frac{u^2}{8g} \] 
Step 4: {Calculate angular momentum} 
\[ L = m u \cos 30^\circ \times H \] \[ = m u \times \frac{\sqrt{3}}{2} \times \frac{u^2}{8g} \] \[ = \frac{\sqrt{3} m u^3}{16 g} \] Thus, the correct answer is (A) \( \frac{\sqrt{3}}{16} \frac{m u^3}{g} \).