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

Explanation

To find the magnitude and direction of the angular momentum of a particle projected from a point P at an angle of \(45^\circ\) to the horizontal, we can follow these steps:

### Step 1: Determine the Initial Velocity Components
The initial speed of the particle is \(v_0\) and it is projected at an angle of \(45^\circ\). We can resolve this velocity into its horizontal and vertical components.

\[ v_{0x} = v_0 \cos(45^\circ) = \frac{v_0}{\sqrt{2}} \]
\[ v_{0y} = v_0 \sin(45^\circ) = \frac{v_0}{\sqrt{2}} \]

### Step 2: Calculate the Position of the Particle at Time \(t = \frac{v_0}{g}\)
At time \(t = \frac{v_0}{g}\), we can find the position of the particle in the x and y directions using the equations of motion.

**For the x-coordinate:**
Since there is no acceleration in the x-direction, the position is given by:
\[ x = v_{0x} \cdot t = \frac{v_0}{\sqrt{2}} \cdot \frac{v_0}{g} = \frac{v_0^2}{\sqrt{2}g} \]

**For the y-coordinate:**
Using the second equation of motion:
\[ y = v_{0y} \cdot t - \frac{1}{2} g t^2 \]
Substituting the values:
\[ y = \frac{v_0}{\sqrt{2}} \cdot \frac{v_0}{g} - \frac{1}{2} g \left(\frac{v_0}{g}\right)^2 \]
\[ y = \frac{v_0^2}{\sqrt{2}g} - \frac{1}{2} g \cdot \frac{v_0^2}{g^2} = \frac{v_0^2}{\sqrt{2}g} - \frac{v_0^2}{2g} \]
To combine these, we can express \(\frac{v_0^2}{2g}\) with a common denominator:
\[ y = \frac{v_0^2}{\sqrt{2}g} - \frac{v_0^2 \sqrt{2}}{2\sqrt{2}g} = \frac{v_0^2(2 - \sqrt{2})}{2\sqrt{2}g} \]

### Step 3: Calculate the Angular Momentum
The angular momentum \(L\) about point P is given by:
\[ L = r \times p \]
where \(r\) is the position vector and \(p\) is the linear momentum.

The linear momentum \(p\) is:
\[ p = m \cdot v \]
The velocity vector at time \(t = \frac{v_0}{g}\) has the same x-component and a new y-component due to gravity:
\[ v_y = v_{0y} - g t = \frac{v_0}{\sqrt{2}} - g \cdot \frac{v_0}{g} = \frac{v_0}{\sqrt{2}} - v_0 = v_0 \left(\frac{1}{\sqrt{2}} - 1\right) \]
Thus, the velocity vector is:
\[ \vec{v} = \left(\frac{v_0}{\sqrt{2}}, v_0 \left(\frac{1}{\sqrt{2}} - 1\right)\right) \]

### Step 4: Calculate the Angular Momentum Components
The position vector \(r\) is:
\[ r = \left(\frac{v_0^2}{\sqrt{2}g}, \frac{v_0^2(2 - \sqrt{2})}{2\sqrt{2}g}\right) \]
The angular momentum \(L\) can be calculated using the determinant:
\[ L = m \left( r_x v_y - r_y v_x \right) \]
Substituting the values:
\[ L = m \left( \frac{v_0^2}{\sqrt{2}g} \cdot v_0 \left(\frac{1}{\sqrt{2}} - 1\right) - \frac{v_0^2(2 - \sqrt{2})}{2\sqrt{2}g} \cdot \frac{v_0}{\sqrt{2}} \right) \]
After simplifying, we find:
\[ L = \frac{mv_0^3}{2g} \left( \sqrt{2} - 1 \right) \]

### Step 5: Determine the Direction of Angular Momentum
Using the right-hand rule, the direction of angular momentum will be negative along the z-axis (downward) since the motion is clockwise.

### Final Result
The magnitude of the angular momentum is:
\[ L = \frac{mv_0^3}{2g} \left( \sqrt{2} - 1 \right) \]
And the direction is:
\[ \text{Direction: } -\hat{k} \text{ (downward)} \]