A small ball is suspended from point O by a thread of length l. A nail is driven into the wall at a distance of `l//2` below O, at A. The ball is drawn aside so that the thread takes up a horizontal position at the level of point O and then released. Find
a. At what angle from the vertical of the ball's trajectory, will the tension in the thread disappear?
b. What will be the highest point from the lowermost point of circular track, to which it will rise?

Explanation

To solve the problem step by step, we will break it down into two parts as specified in the question.

### Part (a): Finding the angle from the vertical at which the tension in the thread disappears.

1. **Understanding the Setup**:
- A ball is suspended from point O by a thread of length \( l \).
- A nail is driven into the wall at point A, which is at a distance \( \frac{l}{2} \) below O.
- The ball is drawn to a horizontal position and released.

2. **Analyzing Forces at Point P**:
- When the ball swings down, it will follow a circular path until it reaches point P, where the tension in the thread becomes zero.
- At point P, the only forces acting on the ball are its weight \( mg \) acting downwards and the component of this weight that provides the centripetal force.

3. **Setting Up the Equation**:
- At point P, the centripetal force required for circular motion is provided by the component of the gravitational force acting along the direction of the radius.
- The equation can be written as:
\[ mg \cos(\theta) = \frac{mv_P^2}{r} \]
- Here, \( r = \frac{l}{2} \) (the radius of the circular path), and \( v_P \) is the velocity at point P.

4. **Using Conservation of Energy**:
- The potential energy lost when the ball moves from the horizontal position to point P is converted into kinetic energy.
- The change in potential energy is:
\[ \Delta PE = mg \left(\frac{l}{2} (1 - \cos(\theta))\right) \]
- The kinetic energy at point P is:
\[ KE = \frac{1}{2} mv_P^2 \]
- Setting these equal gives:
\[ mg \left(\frac{l}{2} (1 - \cos(\theta))\right) = \frac{1}{2} mv_P^2 \]

5. **Substituting \( v_P^2 \)**:
- From the centripetal force equation, we have:
\[ v_P^2 = \frac{g l \cos(\theta)}{2} \]
- Substituting this into the energy equation:
\[ mg \left(\frac{l}{2} (1 - \cos(\theta))\right) = \frac{1}{2} m \left(\frac{g l \cos(\theta)}{2}\right) \]
- Simplifying gives:
\[ g \left(\frac{l}{2} (1 - \cos(\theta))\right) = \frac{g l \cos(\theta)}{4} \]

6. **Solving for \( \cos(\theta) \)**:
- Canceling \( g \) and \( l \) (assuming they are non-zero):
\[ 2(1 - \cos(\theta)) = \cos(\theta) \]
- Rearranging gives:
\[ 2 - 2\cos(\theta) = \cos(\theta) \implies 2 = 3\cos(\theta) \implies \cos(\theta) = \frac{2}{3} \]

7. **Final Result for Part (a)**:
- Therefore, the angle \( \theta \) is:
\[ \theta = \cos^{-1}\left(\frac{2}{3}\right) \]

### Part (b): Finding the highest point from the lowermost point of the circular track.

1. **Finding the Velocity at Point P**:
- We already found that:
\[ v_P^2 = \frac{g l \cos(\theta)}{2} \]
- Substituting \( \cos(\theta) = \frac{2}{3} \):
\[ v_P^2 = \frac{g l \cdot \frac{2}{3}}{2} = \frac{g l}{3} \]
- Thus, \( v_P = \sqrt{\frac{g l}{3}} \).

2. **Projectile Motion After Point P**:
- After reaching point P, the ball will move in projectile motion.
- The maximum height \( h \) reached in projectile motion can be calculated using:
\[ h = \frac{v_P^2 \sin^2(\theta)}{2g} \]

3. **Finding \( \sin(\theta) \)**:
- Using \( \sin^2(\theta) + \cos^2(\theta) = 1 \):
\[ \sin^2(\theta) = 1 - \left(\frac{2}{3}\right)^2 = 1 - \frac{4}{9} = \frac{5}{9} \]

4. **Calculating Maximum Height**:
- Substituting \( v_P \) and \( \sin^2(\theta) \):
\[ h = \frac{\left(\frac{g l}{3}\right) \cdot \frac{5}{9}}{2g} = \frac{5l}{54} \]

5. **Total Height from the Lowermost Point**:
- The total height from the lowest point of the circular track is the height gained plus the distance from point A to the lowest point:
\[ H = \frac{l}{2} + \frac{5l}{54} = \frac{27l}{54} + \frac{5l}{54} = \frac{32l}{54} = \frac{16l}{27} \]

### Final Results:
- **Part (a)**: \( \theta = \cos^{-1}\left(\frac{2}{3}\right) \)
- **Part (b)**: The highest point from the lowermost point of the circular track is \( \frac{16l}{27} \).