A heavy particle hangs from a point O, by a string of length a. It is projected horizontally with a velocity `u = sqrt((2 + sqrt(3))ag)` . The angle with the downward vertical, string makes where string becomes slack is :

Explanation

To solve the problem step by step, we need to analyze the motion of the heavy particle and the forces acting on it when the string becomes slack. Here’s how we can approach this:

### Step 1: Understand the Initial Conditions
The heavy particle is hanging from a point O by a string of length \( a \) and is projected horizontally with a velocity \( u = \sqrt{(2 + \sqrt{3})ag} \).

### Step 2: Draw the Diagram
Draw a diagram showing the point O, the string of length \( a \), and the particle at an angle \( \theta \) from the vertical when the string becomes slack.

### Step 3: Identify Forces Acting on the Particle
At the position where the string becomes slack, the forces acting on the particle are:
- The gravitational force \( mg \) acting downwards.
- The tension \( T \) in the string acting along the string.

### Step 4: Resolve Forces
Resolve the gravitational force into two components:
- \( mg \cos \theta \) (along the string)
- \( mg \sin \theta \) (perpendicular to the string)

### Step 5: Apply the Centripetal Force Condition
At the moment the string becomes slack, the tension \( T \) becomes zero. Therefore, the centripetal force required for circular motion is provided solely by the component of the gravitational force:
\[ T - mg \cos \theta = \frac{mv^2}{a} \]
Since \( T = 0 \) when the string becomes slack, we have:
\[ -mg \cos \theta = \frac{mv^2}{a} \]

### Step 6: Apply the Work-Energy Theorem
Using the work-energy theorem, the initial kinetic energy plus the initial potential energy equals the final kinetic energy plus the final potential energy:
\[ \frac{1}{2}mu^2 + 0 = \frac{1}{2}mv^2 + mg(a(1 - \cos \theta)) \]
Cancel \( m \) from the equation:
\[ \frac{1}{2}u^2 = \frac{1}{2}v^2 + g(a(1 - \cos \theta)) \]

### Step 7: Substitute Values
Substituting \( u = \sqrt{(2 + \sqrt{3})ag} \):
\[ \frac{1}{2}(2 + \sqrt{3})ag = \frac{1}{2}v^2 + g(a(1 - \cos \theta)) \]
Rearranging gives:
\[ (2 + \sqrt{3})ag = v^2 + 2g(a(1 - \cos \theta)) \]

### Step 8: Solve for \( v^2 \)
From the centripetal force equation:
\[ 0 = -mg \cos \theta + \frac{mv^2}{a} \]
This implies:
\[ v^2 = ag \cos \theta \]

### Step 9: Substitute \( v^2 \) in Work-Energy Equation
Substituting \( v^2 = ag \cos \theta \) into the work-energy equation:
\[ (2 + \sqrt{3})ag = ag \cos \theta + 2g(a(1 - \cos \theta)) \]
Simplifying gives:
\[ (2 + \sqrt{3}) = \cos \theta + 2(1 - \cos \theta) \]
\[ (2 + \sqrt{3}) = 2 - \cos \theta \]
\[ \cos \theta = 2 - (2 + \sqrt{3}) = -\sqrt{3} \]

### Step 10: Find \( \theta \)
Thus, we have:
\[ \cos \theta = -\frac{1}{\sqrt{3}} \]
Therefore, the angle \( \theta \) is:
\[ \theta = \cos^{-1}(-\frac{1}{\sqrt{3}}) \]

### Final Answer
The angle with the downward vertical that the string makes when it becomes slack is:
\[ \theta = \cos^{-1}(-\frac{1}{\sqrt{3}}) \]

---