A particle of mass m is projected with a velocity v making an angle of `45^@` with the horizontal. The magnitude of the angular momentum of the projectile abut the point of projection when the particle is at its maximum height h is.

Options

A. zero
B. `(mv)^3/(4sqrt2g)`
C. `(mv)^3/(sqrt2g)`
D. `msqrt(2gh^3)`

Explanation

To solve the problem of finding the magnitude of the angular momentum of a projectile about the point of projection when it is at its maximum height, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the particle: \( m \)
- Initial velocity of the particle: \( v \)
- Angle of projection: \( \theta = 45^\circ \)

2. **Determine the Components of Velocity:**
- The horizontal component of the velocity \( v_x \) at the maximum height is given by:
\[ v_x = v \cos(\theta) = v \cos(45^\circ) = \frac{v}{\sqrt{2}} \]
- The vertical component of the velocity \( v_y \) at the maximum height is:
\[ v_y = v \sin(\theta) = v \sin(45^\circ) = \frac{v}{\sqrt{2}} \]
- At the maximum height, \( v_y = 0 \).

3. **Calculate the Maximum Height \( h \):**
- The formula for maximum height \( h \) in projectile motion is:
\[ h = \frac{v^2 \sin^2(\theta)}{2g} \]
- Substituting \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ h = \frac{v^2 \left(\frac{1}{\sqrt{2}}\right)^2}{2g} = \frac{v^2 \cdot \frac{1}{2}}{2g} = \frac{v^2}{4g} \]

4. **Determine the Angular Momentum \( L \):**
- The angular momentum \( L \) about the point of projection is given by:
\[ L = m \cdot v \cdot r_{\perp} \]
- Here, \( r_{\perp} \) is the perpendicular distance from the line of motion to the point of projection, which at maximum height is equal to the maximum height \( h \).
- Thus, we have:
\[ L = m \cdot v_x \cdot h \]
- Substituting \( v_x = \frac{v}{\sqrt{2}} \) and \( h = \frac{v^2}{4g} \):
\[ L = m \cdot \left(\frac{v}{\sqrt{2}}\right) \cdot \left(\frac{v^2}{4g}\right) = \frac{m v^3}{4g \sqrt{2}} \]

5. **Final Expression for Angular Momentum:**
- Therefore, the magnitude of the angular momentum of the projectile about the point of projection when the particle is at its maximum height \( h \) is:
\[ L = \frac{m v^3}{4g \sqrt{2}} \]