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 step by step, we will follow the trajectory of the projectile and determine the radius of curvature when the velocity makes an angle of \(30^\circ\) with the horizontal.

### Step 1: Identify the initial conditions
The particle is projected with an initial speed \(u = 20 \, \text{m/s}\) at an angle of \(\theta = 60^\circ\) with the horizontal.

### Step 2: Calculate the components of the initial velocity
The horizontal and vertical components of the initial velocity can be calculated as:
- \(u_x = u \cos(\theta) = 20 \cos(60^\circ) = 20 \times \frac{1}{2} = 10 \, \text{m/s}\)
- \(u_y = u \sin(\theta) = 20 \sin(60^\circ) = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{m/s}\)

### Step 3: Determine the velocity at the point where the angle is \(30^\circ\)
At the point where the velocity makes an angle of \(30^\circ\) with the horizontal, we can denote the components of the velocity as \(V_x\) and \(V_y\):
- From the angle, we know that \(\tan(30^\circ) = \frac{V_y}{V_x}\) implies \(V_y = V_x \tan(30^\circ) = V_x \cdot \frac{1}{\sqrt{3}}\).

### Step 4: Relate the vertical and horizontal components of velocity
Using the horizontal component:
- \(V_x = u_x = 10 \, \text{m/s}\) (since horizontal velocity remains constant)
- Therefore, \(V_y = 10 \cdot \frac{1}{\sqrt{3}} = \frac{10}{\sqrt{3}} \, \text{m/s}\).

### Step 5: Calculate the net velocity at this point
The net velocity \(V\) can be calculated using the Pythagorean theorem:
\[ V = \sqrt{V_x^2 + V_y^2} = \sqrt{(10)^2 + \left(\frac{10}{\sqrt{3}}\right)^2} = \sqrt{100 + \frac{100}{3}} = \sqrt{\frac{300 + 100}{3}} = \sqrt{\frac{400}{3}} = \frac{20}{\sqrt{3}} \, \text{m/s}. \]

### Step 6: Calculate the radius of curvature
The radius of curvature \(R\) for projectile motion can be given by the formula:
\[ R = \frac{V^2}{g \cos(\theta)}, \]
where \(g = 10 \, \text{m/s}^2\) and \(\theta = 30^\circ\) (the angle of the velocity with the horizontal).

Substituting the values:
\[ R = \frac{\left(\frac{20}{\sqrt{3}}\right)^2}{10 \cdot \cos(30^\circ)} = \frac{\frac{400}{3}}{10 \cdot \frac{\sqrt{3}}{2}} = \frac{\frac{400}{3}}{5\sqrt{3}} = \frac{80}{3\sqrt{3}} \, \text{m}. \]

### Step 7: Set the radius equal to the given expression
According to the problem, the radius of curvature is also given as:
\[ R = \frac{80}{9\sqrt{x}}. \]
Setting these equal gives:
\[ \frac{80}{3\sqrt{3}} = \frac{80}{9\sqrt{x}}. \]

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

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

---