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`

Options

A. a. `(mv_(0)^(3))/(2sqrt(2)g)`
B. b. `(mv_(0)^(3))/(sqrt(2)g)`
C. c. `(3mv_(0)^(3))/(sqrt(2)g)`
D. d. `(sqrt(2)mv_(0)^(3))/(g)`

Explanation

To solve the problem of finding the magnitude and direction of the angular momentum of a particle projected at an angle of \(45^\circ\) with an initial speed \(v_0\) at time \(t = \frac{v_0}{g}\), we can follow these steps:

### Step 1: Determine the initial velocity components
The initial velocity \(v_0\) can be resolved into its horizontal and vertical components using trigonometric functions.

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

### Step 2: Calculate the velocity at time \(t = \frac{v_0}{g}\)
The horizontal component of velocity remains constant throughout the motion, while the vertical component changes due to gravity.

\[ v_x = v_{0x} = \frac{v_0}{\sqrt{2}} \]
\[ v_y = v_{0y} - g t = \frac{v_0}{\sqrt{2}} - g \left(\frac{v_0}{g}\right) = \frac{v_0}{\sqrt{2}} - v_0 = \frac{v_0(1 - \sqrt{2})}{\sqrt{2}} \]

Thus, the velocity vector at time \(t = \frac{v_0}{g}\) is:

\[ \vec{v} = \left(\frac{v_0}{\sqrt{2}}\right) \hat{i} + \left(\frac{v_0(1 - \sqrt{2})}{\sqrt{2}}\right) \hat{j} \]

### Step 3: Calculate the position at time \(t = \frac{v_0}{g}\)
The position vector can be calculated using the equations of motion.

\[ x = v_{0x} \cdot t = \left(\frac{v_0}{\sqrt{2}}\right) \left(\frac{v_0}{g}\right) = \frac{v_0^2}{\sqrt{2}g} \]
\[ y = v_{0y} \cdot t - \frac{1}{2} g t^2 = \left(\frac{v_0}{\sqrt{2}}\right) \left(\frac{v_0}{g}\right) - \frac{1}{2} g \left(\frac{v_0}{g}\right)^2 \]
\[ = \frac{v_0^2}{\sqrt{2}g} - \frac{1}{2} \cdot \frac{v_0^2}{g} = \frac{v_0^2}{\sqrt{2}g} - \frac{v_0^2}{2g} = \frac{v_0^2(2 - \sqrt{2})}{2\sqrt{2}g} \]

Thus, the position vector at time \(t = \frac{v_0}{g}\) is:

\[ \vec{r} = \left(\frac{v_0^2}{\sqrt{2}g}\right) \hat{i} + \left(\frac{v_0^2(2 - \sqrt{2})}{2\sqrt{2}g}\right) \hat{j} \]

### Step 4: Calculate the angular momentum
The angular momentum \(\vec{L}\) about point \(P\) is given by:

\[ \vec{L} = m \vec{r} \times \vec{v} \]

Calculating the cross product, we have:

\[ \vec{L} = m \left( \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{v_0^2}{\sqrt{2}g} & \frac{v_0^2(2 - \sqrt{2})}{2\sqrt{2}g} & 0 \\ \frac{v_0}{\sqrt{2}} & \frac{v_0(1 - \sqrt{2})}{\sqrt{2}} & 0 \end{vmatrix} \right) \]

Calculating the determinant, we find:

\[ L_z = m \left( \frac{v_0^2}{\sqrt{2}g} \cdot \frac{v_0(1 - \sqrt{2})}{\sqrt{2}} - \frac{v_0^2(2 - \sqrt{2})}{2\sqrt{2}g} \cdot \frac{v_0}{\sqrt{2}} \right) \]

Simplifying gives:

\[ L_z = m \left( \frac{v_0^3(1 - \sqrt{2})}{2g} - \frac{v_0^3(2 - \sqrt{2})}{4g} \right) = m \frac{v_0^3}{2g} \left(1 - \sqrt{2} - \frac{(2 - \sqrt{2})}{2}\right) \]

This results in:

\[ L = -\frac{m v_0^3}{2\sqrt{2}g} \hat{k} \]

### Step 5: Conclusion
The magnitude of the angular momentum is:

\[ |\vec{L}| = \frac{m v_0^3}{2\sqrt{2}g} \]

And the direction is in the negative \(z\)-direction (or \(-\hat{k}\)).