A cylindrical hall has a horizontal smooth floor. A ball is projected along the floor from A point on the wall in a direction making an angle `theta` with the radius through the point the ball returns bak to the initial point after two impacts with the wall. if the coefficient of restitution is e than `tan^(2)theta` will be

Options

A. `(1+e+e^(2))/(e^(3))`
B. `(1+e)/(e^(2))`
C. `(e^(2))/(1+e)`
D. `(e^(3))/(1+e+e^(2))`

Explanation

To solve the problem, we need to analyze the motion of the ball as it collides with the walls of the cylindrical hall. Here’s the step-by-step solution:

### Step 1: Understanding the Setup
The ball is projected from point A on the wall at an angle \( \theta \) with respect to the radius through point A. The ball will collide with the wall twice before returning to point A.

### Step 2: Analyzing the First Collision
1. **Initial Velocity Components**: When the ball is projected, it has two components of velocity:
- Tangential component: \( u \sin \theta \)
- Normal component: \( u \cos \theta \)

2. **Collision with the Wall**: Upon colliding with the wall, the normal component of the velocity will change due to the coefficient of restitution \( e \). The velocity after the first collision can be expressed as:
\[ V_1 \cos \alpha = e \cdot (u \cos \theta) \]
Here, \( V_1 \) is the velocity after the first collision, and \( \alpha \) is the angle of incidence.

### Step 3: Analyzing the Second Collision
1. **Velocity Components After First Collision**: After the first collision, the tangential component remains unchanged:
\[ V_1 \sin \alpha = u \sin \theta \]

2. **Second Collision**: After the second collision, the same logic applies:
\[ V_2 \cos \beta = e \cdot V_1 \cos \alpha \]
The tangential component remains the same:
\[ V_2 \sin \beta = V_1 \sin \alpha \]

### Step 4: Relating Angles
1. **Angle Relationships**: From the geometry of the problem, we can establish:
\[ 2\alpha + \beta + \theta = 180^\circ \]
This implies:
\[ \alpha + \beta = 90^\circ - \theta \]

### Step 5: Using Trigonometric Identities
1. **Tangent Relationships**: Taking the tangent of both sides gives:
\[ \tan(90^\circ - \theta) = \cot(\theta) = \frac{1}{\tan(\theta)} \]
Thus:
\[ \tan \alpha + \tan \beta = \frac{1}{\tan \theta} \]

2. **Substituting Values**: We can express \( \tan \beta \) in terms of \( \tan \alpha \):
\[ \tan \beta = \frac{\tan \alpha}{e} \]
Substituting this into the earlier equation gives:
\[ \tan \alpha + \frac{\tan \alpha}{e} = \frac{1}{\tan \theta} \]

### Step 6: Solving for \( \tan^2 \theta \)
1. **Combining Terms**: Factoring out \( \tan \alpha \):
\[ \tan \alpha \left(1 + \frac{1}{e}\right) = \frac{1}{\tan \theta} \]
Thus:
\[ \tan \alpha = \frac{1}{\tan \theta \left(1 + \frac{1}{e}\right)} \]

2. **Final Expression**: After substituting and rearranging, we find:
\[ \tan^2 \theta = \frac{e^3}{1 + e + e^2} \]

### Final Answer
Thus, the expression for \( \tan^2 \theta \) is:
\[ \tan^2 \theta = \frac{e^3}{1 + e + e^2} \]