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))/(4sqrt(2g))`
C. `(mV^(3))/sqrt(2g)`
D. `(mV)/(sqrt(2gh^(3)))`

Explanation

To find 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 1: Determine the maximum height (h)
The maximum height (h) reached by a projectile is given by the formula:
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]
where \( u \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

Given that \( u = v \) and \( \theta = 45^\circ \), we can calculate:
\[ h = \frac{v^2 \sin^2(45^\circ)}{2g} \]
Since \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ h = \frac{v^2 \left(\frac{1}{\sqrt{2}}\right)^2}{2g} = \frac{v^2 \cdot \frac{1}{2}}{2g} = \frac{v^2}{4g} \]

### Step 2: Find the horizontal component of velocity at maximum height
At maximum height, the vertical component of the velocity becomes zero, and only the horizontal component remains. The horizontal component of the initial velocity is:
\[ v_x = u \cos \theta = v \cos(45^\circ) = v \cdot \frac{1}{\sqrt{2}} = \frac{v}{\sqrt{2}} \]

### Step 3: Calculate the linear momentum (p)
The linear momentum (p) of the particle at maximum height is given by:
\[ p = m v_x = m \cdot \frac{v}{\sqrt{2}} = \frac{mv}{\sqrt{2}} \]

### Step 4: Determine the perpendicular distance from the line of motion to the point of projection
At maximum height, the perpendicular distance from the line of motion (which is horizontal) to the point of projection is equal to the maximum height \( h \):
\[ d = h = \frac{v^2}{4g} \]

### Step 5: Calculate the angular momentum (L)
The angular momentum (L) about the point of projection is given by:
\[ L = p \cdot d \]
Substituting the values we have:
\[ L = \left(\frac{mv}{\sqrt{2}}\right) \cdot \left(\frac{v^2}{4g}\right) \]
\[ L = \frac{mv \cdot v^2}{4g \sqrt{2}} = \frac{mv^3}{4g \sqrt{2}} \]

### Final Answer
The magnitude of the angular momentum of the projectile about the point of projection when the particle is at its maximum height \( h \) is:
\[ L = \frac{mv^3}{4g \sqrt{2}} \]

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