Solution:

The angular momentum of a projectile about the point of projection when it reaches its maximum height is given by:

$$L = m v_x r$$

where:

• 
  $m$ 

  = mass of the projectile

• 
  $v_x$ 

  = horizontal component of velocity

• 
  $r$ 

  = perpendicular distance from the point of projection (which is the maximum height $h$ 

  )

Step 1: Horizontal Component of Velocity

The initial velocity components are:

$$v_x = v \cos 45^\circ = \frac{v}{\sqrt{2}}$$

$$v_y = v \sin 45^\circ = \frac{v}{\sqrt{2}}$$

Since there is no acceleration in the horizontal direction (ignoring air resistance),

$v_x$

remains constant throughout the motion.

Step 2: Maximum Height

Using the kinematic equation:

$$v_y^2 = u_y^2 - 2 g h$$

At maximum height, the vertical velocity becomes zero, so:

$$0 = \left(\frac{v}{\sqrt{2}}\right)^2 - 2 g h$$

Solving for

$h$

:

$$h = \frac{v^2}{2 g} \cdot \frac{1}{2} = \frac{v^2}{4g}$$

Step 3: Angular Momentum Calculation

The angular momentum at maximum height is:

$$L = m v_x h$$

Substituting values:

$$L = m \left(\frac{v}{\sqrt{2}}\right) \left(\frac{v^2}{4g}\right)$$

$$L = m \cdot \frac{v}{\sqrt{2}} \cdot \frac{v^2}{4g}$$

$$L = \frac{m v^3}{4g \sqrt{2}}$$

Thus, the magnitude of the angular momentum about the point of projection at maximum height is:

$$\frac{m v^3}{4g \sqrt{2}}$$