A body is projected at an angle of `30^(@)` with the horizontal with momentum `P`.At its highest point the magnitude of the momentum is:

Options

A. `(sqrt(3))/(2)p`
B. `(2)/(sqrt(3))p`
C. p
D. `(p)/(2)`

Explanation

To solve the problem, we need to analyze the motion of the body projected at an angle of \(30^\circ\) with an initial momentum \(P\). We will find the magnitude of the momentum at the highest point of its trajectory.

### Step-by-Step Solution:

1. **Understanding the Initial Momentum**:
The initial momentum \(P\) of the body can be expressed as:
\[ P = m \cdot V_0 \]
where \(m\) is the mass of the body and \(V_0\) is the initial velocity of projection.

2. **Components of Initial Velocity**:
The initial velocity \(V_0\) can be broken down into its horizontal and vertical components:
- Horizontal component:
\[ V_{0x} = V_0 \cos(30^\circ) = V_0 \cdot \frac{\sqrt{3}}{2} \]
- Vertical component:
\[ V_{0y} = V_0 \sin(30^\circ) = V_0 \cdot \frac{1}{2} \]

3. **Velocity at the Highest Point**:
At the highest point of the projectile's trajectory, the vertical component of the velocity becomes zero. Therefore, the velocity at the highest point \(V_A\) is only the horizontal component:
\[ V_A = V_{0x} = V_0 \cdot \frac{\sqrt{3}}{2} \]

4. **Calculating Momentum at the Highest Point**:
The momentum at the highest point \(P_A\) can be calculated using the mass and the horizontal velocity:
\[ P_A = m \cdot V_A = m \cdot \left(V_0 \cdot \frac{\sqrt{3}}{2}\right) \]

5. **Relating \(P_A\) to Initial Momentum \(P\)**:
Since we know that \(P = m \cdot V_0\), we can substitute this into our equation for \(P_A\):
\[ P_A = m \cdot V_0 \cdot \frac{\sqrt{3}}{2} = P \cdot \frac{\sqrt{3}}{2} \]

6. **Final Expression for Momentum at the Highest Point**:
Thus, the magnitude of the momentum at the highest point is:
\[ P_A = \frac{\sqrt{3}}{2} P \]

### Conclusion:
The magnitude of the momentum at the highest point is \(\frac{\sqrt{3}}{2} P\).