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) \( \frac{\sqrt{3}}{16} \frac{m u^3}{g} \). 
 

Answer 2

Given:
- Mass of particle = \( m \)
- Initial velocity = \( u \)
- Angle of projection = \( 30^\circ \)
We are to find the **magnitude of angular momentum** of the particle **about the point of projection** when it is at its **maximum height**.

Step 1: Concept of angular momentum at max height
At maximum height:
- The vertical velocity becomes 0
- The horizontal component of velocity remains \( u \cos 30^\circ \)
- The particle is at a horizontal distance \( R/2 \) from the point of projection
- Height = \( h \)
The angular momentum \( L \) about the point of projection is:
\[ L = m \cdot v \cdot r \cdot \sin\theta \]
Where:
- \( v = \) horizontal velocity = \( u \cos 30^\circ \)
- \( r = \) perpendicular distance from origin = height \( h \)
- \( \sin\theta = 1 \) because velocity is perpendicular to radius vector at maximum height

So:
\[ L = m u \cos 30^\circ \cdot h = m u \cdot \frac{\sqrt{3}}{2} \cdot h \]

Step 2: Find height \( h \)
Maximum height is given by:
\[ h = \frac{u^2 \sin^2\theta}{2g} = \frac{u^2 \sin^2 30^\circ}{2g} = \frac{u^2 \cdot (1/2)^2}{2g} = \frac{u^2}{8g} \]

Step 3: Substitute \( h \) into angular momentum expression
\[ L = m u \cdot \frac{\sqrt{3}}{2} \cdot \frac{u^2}{8g} = \frac{\sqrt{3}}{2} \cdot \frac{m u^3}{8g} = \frac{\sqrt{3}}{16} \cdot \frac{m u^3}{g} \]

Final Answer:
The angular momentum at maximum height is:
\[ \boxed{\frac{\sqrt{3}}{16} \cdot \frac{m u^3}{g}} \]