A particle is projected at time t=0 from a point P wilth a speed `v_0` at an angle of `45^@` to the horizontal. Find the magnitude and the direction of the angular momentum of the particle about the point P at time `t=v_0/g`

Explanation

To find the magnitude and direction of the angular momentum of a particle projected from point P with speed \( v_0 \) at an angle of \( 45^\circ \) to the horizontal at time \( t = \frac{v_0}{g} \), we can follow these steps:

### Step 1: Determine the Components of Initial Velocity
The initial velocity \( v_0 \) can be broken down into its horizontal and vertical components:
- \( u_x = v_0 \cos(45^\circ) = v_0 \cdot \frac{1}{\sqrt{2}} = \frac{v_0}{\sqrt{2}} \)
- \( u_y = v_0 \sin(45^\circ) = v_0 \cdot \frac{1}{\sqrt{2}} = \frac{v_0}{\sqrt{2}} \)

### Step 2: Find Position of the Particle at \( t = \frac{v_0}{g} \)
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