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 will analyze the motion of the heavy particle and apply the principles of energy conservation and forces acting on the particle.

### Step 1: Understand the Problem
A heavy particle is hanging from a point O by a string of length \( a \) and is projected horizontally with a velocity \( u = \sqrt{(2 + \sqrt{3})ag} \). We need to find the angle \( \theta \) with the downward vertical at which the string becomes slack.

### Step 2: Identify Forces Acting on the Particle
When the particle is in motion, the forces acting on it are:
- The gravitational force \( mg \) acting downwards.
- The tension \( T \) in the string acting along the string towards point O.

For the string to become slack, the tension \( T \) must be zero.

### Step 3: Apply the Condition for Slack
At the point where the string becomes slack, we can write the equation of motion in terms of the forces acting on the particle. The centripetal force required for circular motion is provided by the component of the gravitational force and the tension in the string.

Using the radial force balance:
\[ T - mg \cos \theta = \frac{mv^2}{a} \]
Setting \( T = 0 \) (since the string becomes slack), we have:
\[ -mg \cos \theta = \frac{mv^2}{a} \]
This simplifies to:
\[ g \cos \theta = \frac{v^2}{a} \]

### Step 4: Apply the Work-Energy Theorem
We will use the work-energy principle to relate the initial kinetic energy and potential energy to the final kinetic energy and potential energy.

The initial kinetic energy \( K_i \) when the particle is projected is:
\[ K_i = \frac{1}{2} m u^2 = \frac{1}{2} m (2 + \sqrt{3})ag \]
The initial potential energy \( U_i \) is zero (taking the reference level at the height of the point O).

The final kinetic energy \( K_f \) when the particle has moved and is at angle \( \theta \) is:
\[ K_f = \frac{1}{2} mv^2 \]
The potential energy \( U_f \) at angle \( \theta \) is:
\[ U_f = mg(a - a \cos \theta) = mga(1 - \cos \theta) \]

By conservation of mechanical energy:
\[ K_i + U_i = K_f + U_f \]
Substituting the values:
\[ \frac{1}{2} m (2 + \sqrt{3})ag + 0 = \frac{1}{2} mv^2 + mga(1 - \cos \theta) \]

### Step 5: Simplify and Solve for \( v^2 \)
This gives:
\[ \frac{1}{2} (2 + \sqrt{3})ag = \frac{1}{2} v^2 + g a (1 - \cos \theta) \]
Multiplying through by 2:
\[ (2 + \sqrt{3})ag = v^2 + 2ga(1 - \cos \theta) \]
Rearranging gives:
\[ v^2 = (2 + \sqrt{3})ag - 2ga(1 - \cos \theta) = (2 + \sqrt{3} - 2 + 2\cos \theta)ag = (\sqrt{3} + 2\cos \theta)ag \]

### Step 6: Substitute \( v^2 \) into the Radial Force Equation
Now substitute \( v^2 \) back into the radial force equation:
\[ g \cos \theta = \frac{(\sqrt{3} + 2\cos \theta)ag}{a} \]
This simplifies to:
\[ g \cos \theta = g(\sqrt{3} + 2\cos \theta) \]
Dividing both sides by \( g \) (assuming \( g \neq 0 \)):
\[ \cos \theta = \sqrt{3} + 2\cos \theta \]
Rearranging gives:
\[ -\cos \theta = \sqrt{3} \implies \cos \theta = -\frac{1}{\sqrt{3}} \]

### Step 7: Find the Angle \( \theta \)
Thus, we have:
\[ \theta = \cos^{-1}\left(-\frac{1}{\sqrt{3}}\right) \]

### Final Answer
The angle \( \theta \) with the downward vertical at which the string becomes slack is:
\[ \theta = \cos^{-1}\left(-\frac{1}{\sqrt{3}}\right) \]