A particle is projected from the ground at an angle of `60^@` with the horizontal with a speed by 20 m/s. The radius of curvature of the path of the particle, when its velocity makes an angle of `30^@` with horizontal is `80/9.sqrt(x)`m Find x .

Explanation

To solve the problem, we need to find the value of \( x \) in the expression for the radius of curvature of the projectile's path when its velocity makes an angle of \( 30^\circ \) with the horizontal.

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- The particle is projected from the ground at an angle of \( 60^\circ \) with an initial speed of \( 20 \, \text{m/s} \).

2. **Determine the Components of Initial Velocity**:
- The horizontal component of the initial velocity \( V_x \) is given by:
\[ V_x = V \cos(60^\circ) = 20 \cdot \frac{1}{2} = 10 \, \text{m/s} \]
- The vertical component of the initial velocity \( V_y \) is given by:
\[ V_y = V \sin(60^\circ) = 20 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{m/s} \]

3. **Find the Velocity When the Angle is \( 30^\circ \)**:
- At the point where the velocity makes an angle of \( 30^\circ \) with the horizontal, the horizontal component of the velocity \( V_{x'} \) is:
\[ V_{x'} = V \cos(30^\circ) = V \cdot \frac{\sqrt{3}}{2} \]
- The vertical component of the velocity \( V_{y'} \) is:
\[ V_{y'} = V \sin(30^\circ) = V \cdot \frac{1}{2} \]

4. **Use the Relationship of Components**:
- Since the horizontal velocity remains constant throughout the projectile motion:
\[ V_{x'} = 10 \, \text{m/s} \]
- Thus, we can set:
\[ 10 = V \cdot \frac{\sqrt{3}}{2} \implies V = \frac{20}{\sqrt{3}} \, \text{m/s} \]

5. **Determine the Acceleration Components**:
- The only acceleration acting on the projectile is due to gravity, which acts downwards:
\[ A = g = 10 \, \text{m/s}^2 \]
- The component of acceleration perpendicular to the velocity at \( 30^\circ \) is:
\[ A_{\perpendicular} = g \cos(30^\circ) = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s}^2 \]

6. **Calculate the Radius of Curvature (ROC)**:
- The formula for the radius of curvature \( R \) is:
\[ R = \frac{V^2}{A_{\perpendicular}} \]
- Substituting the values:
\[ R = \frac{\left(\frac{20}{\sqrt{3}}\right)^2}{5\sqrt{3}} = \frac{\frac{400}{3}}{5\sqrt{3}} = \frac{400}{15\sqrt{3}} = \frac{80}{3\sqrt{3}} \, \text{m} \]

7. **Equate to Given Expression**:
- We know from the problem statement that:
\[ R = \frac{80}{9\sqrt{x}} \, \text{m} \]
- Setting the two expressions for \( R \) equal:
\[ \frac{80}{3\sqrt{3}} = \frac{80}{9\sqrt{x}} \]

8. **Solve for \( x \)**:
- Cross-multiplying gives:
\[ 80 \cdot 9\sqrt{x} = 80 \cdot 3\sqrt{3} \]
- Simplifying:
\[ 9\sqrt{x} = 3\sqrt{3} \implies \sqrt{x} = \frac{3\sqrt{3}}{9} = \frac{\sqrt{3}}{3} \]
- Squaring both sides:
\[ x = \frac{3}{9} = \frac{1}{3} \]

### Final Answer:
Thus, the value of \( x \) is \( 3 \).