A ball is thrown at an angle of incidence `'theta'` on a horizontal plane such that the incident direction and the reflected direction are at right angle to each other. If the coefficient of restuitution is `'e'` then `'theta'` is equal to

Options

A. `tan^(-1) ( e)`
B. `tan^(-1) (2e)`
C. `tan^(-1) (sqrt(2) e)`
D. `tan^(-1) (sqrt( e))`

Explanation

To solve the problem, we need to analyze the motion of the ball when it is thrown at an angle `theta` and reflects off a horizontal plane. Given that the angle of incidence and the angle of reflection are at right angles to each other, we can derive the relationship involving the coefficient of restitution `e`.

### Step-by-Step Solution:

1. **Understanding the Angles**:
- Let the angle of incidence be `theta`.
- The angle of reflection will therefore be `90° - theta` because they are complementary (sum to 90 degrees).

2. **Velocity Components**:
- When the ball is thrown with an initial velocity `u`, we can resolve this velocity into two components:
- Horizontal component (along the plane): \( u \cos \theta \)
- Vertical component (perpendicular to the plane): \( u \sin \theta \)

3. **Velocity After Reflection**:
- After the collision with the plane, the vertical component of the velocity will change due to the coefficient of restitution `e`. The new vertical component of the velocity after reflection will be:
- \( v = e \cdot (u \cos \theta) \)
- The horizontal component remains unchanged:
- \( u \sin \theta \)

4. **Using the Right Angle Condition**:
- According to the problem, the incident direction and the reflected direction are at right angles to each other. Therefore, we can set up the following relationship using the tangent function:
\[ \tan(90° - \theta) = \frac{u \sin \theta}{e \cdot (u \cos \theta)} \]
- Simplifying this gives:
\[ \cot(\theta) = \frac{u \sin \theta}{e \cdot (u \cos \theta)} \]
- This simplifies further to:
\[ \cot(\theta) = \frac{\sin \theta}{e \cos \theta} \]

5. **Rearranging the Equation**:
- We can rewrite the cotangent in terms of tangent:
\[ \frac{1}{\tan(\theta)} = \frac{\sin \theta}{e \cos \theta} \]
- Cross-multiplying gives:
\[ e \sin \theta = \cos \theta \tan(\theta) \]

6. **Final Equation**:
- Rearranging leads to:
\[ e = \tan(\theta) \cdot \cot(\theta) \]
- Since \( \tan(\theta) = \frac{\sin \theta}{\cos \theta} \), we can find that:
\[ e = \frac{\sin^2 \theta}{\cos^2 \theta} \]
- Thus, we can conclude that:
\[ e = \tan^2(\theta) \]

7. **Finding `theta`**:
- From the equation \( e = \tan^2(\theta) \), we can find `theta`:
\[ \theta = \tan^{-1}(\sqrt{e}) \]

### Conclusion:
The angle `theta` is given by:
\[ \theta = \tan^{-1}(\sqrt{e}) \]