A particle of mass m is projected with velocity v making an angle of 45° with the horizontal. The magnitude of the angular momentum of the particle about the point of projection when the particle is at its maximum height is (where g = acceleration due to gravity)

The correct option is mv³/ (4 ��(2)g).

To find the magnitude of the angular momentum of the projectile about the point of projection when it is at its maximum height, we can break down the problem step-by-step.

Step 1: Understand the formula for angular momentum
The angular momentum $\overrightarrow{L}$ of a particle of mass $m$ moving with velocity $\overrightarrow{v}$ about a point is given by:
$\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}=m(\overrightarrow{r}\times \overrightarrow{v})$
The magnitude of the angular momentum is:
$L=m\cdot v\cdot r_{\perp }$
where $r_{\perp }$ is the perpendicular distance from the point of projection (origin) to the line of motion of the particle.

Step 2: Find the velocity at maximum height
At the maximum height, the vertical component of the velocity is zero, and the particle has only horizontal velocity.
The horizontal component of the velocity remains constant throughout the motion:
$v_h=v\cos \left({45}^{\circ }\right)$
Since $\cos \left({45}^{\circ }\right)=\frac{1}{\sqrt{2}}$:
$v_h=\frac{v}{\sqrt{2}}$

Step 3: Determine the perpendicular distance
At the maximum height, the particle moves horizontally. The line of motion is horizontal, and its height above the ground is the maximum height $H$ of the projectile.
Therefore, the perpendicular distance from the point of projection to the line of motion is equal to the maximum height:
$r_{\perp }=H$
The formula for the maximum height of a projectile is:
$H=\frac{v^2{\sin }^2\theta }{2g}$
Substituting $\theta ={45}^{\circ }$ and $\sin \left({45}^{\circ }\right)=\frac{1}{\sqrt{2}}$:
$H=\frac{v^2\cdot {\left(\frac{1}{\sqrt{2}}\right)}^2}{2g}=\frac{v^2}{4g}$

Step 4: Calculate the magnitude of angular momentum
Now we substitute the values of the horizontal velocity $v_h$ and the maximum height $H$ into the angular momentum equation:
$L=m\cdot v_h\cdot H$
$L=m\cdot \left(\frac{v}{\sqrt{2}}\right)\cdot \left(\frac{v^2}{4g}\right)$
$L=\frac{m{v}^3}{4\sqrt{2}g}$