A particle of mass m is projected with velocity v moving at an angle of `45^(@)` with horizontal. The magnitude of angular momentum of projectile about point of projection when particle is at maximum height, is

Options

A. zero
B. `(mvh)/(sqrt(2))`
C. `mvh`
D. `sqrt(2) mvh`

Explanation

To find the magnitude of angular momentum of the projectile about the point of projection when the particle is at maximum height, we can follow these steps:

### Step 1: Understand the Components of Velocity
When the particle is projected at an angle of \(45^\circ\) with the horizontal, its initial velocity \(v\) can be broken down into horizontal and vertical components.

- Horizontal component:
\[ v_x = v \cos(45^\circ) = \frac{v}{\sqrt{2}} \]

- Vertical component:
\[ v_y = v \sin(45^\circ) = \frac{v}{\sqrt{2}} \]

### Step 2: Determine the Velocity at Maximum Height
At maximum height, the vertical component of the velocity becomes zero, and only the horizontal component remains. Therefore, the velocity of the projectile at maximum height \(P\) is:
\[ v_P = v_x = \frac{v}{\sqrt{2}} \]

### Step 3: Calculate the Height at Maximum Height
For a projectile launched at an angle, the maximum height \(h\) can be calculated using the formula:
\[ h = \frac{v_y^2}{2g} = \frac{\left(\frac{v}{\sqrt{2}}\right)^2}{2g} = \frac{v^2/2}{2g} = \frac{v^2}{4g} \]

### Step 4: Find the Angular Momentum
The angular momentum \(L\) about the point of projection \(O\) when the particle is at maximum height can be calculated using the formula:
\[ L = m \cdot v_P \cdot h \]
Substituting the values we found:
\[ L = m \cdot \left(\frac{v}{\sqrt{2}}\right) \cdot h \]
Now substituting \(h\):
\[ L = m \cdot \left(\frac{v}{\sqrt{2}}\right) \cdot \left(\frac{v^2}{4g}\right) \]
This simplifies to:
\[ L = \frac{m v^3}{4g \sqrt{2}} \]

### Step 5: Final Expression for Angular Momentum
However, we can express the angular momentum in terms of \(h\) directly:
\[ L = m \cdot \frac{v}{\sqrt{2}} \cdot h \]
Thus, we can write:
\[ L = \frac{m v h}{\sqrt{2}} \]

### Conclusion
The magnitude of angular momentum of the projectile about the point of projection when the particle is at maximum height is:
\[ L = \frac{m v h}{\sqrt{2}} \]