A projectile is projected with some initial velocity and some initial angle of projection. A wind is also blowing, due to which constant horizontal retardation a is imparted to the particle in the plane of motion. It is found that, the particle is at same height at two different time `t_(1)` & `t_(2)` and particle is at same horizontal distance at two different time `t_(3)` & `t_(4)`
Angle of projection of particle is :-

Options

A. `tan^(-1)((t_(1)+t_(2))/(t_(3)+t_(4)))`
B. `tan^(-1)[(a(t_(1)+t_(2)))/(g(t_(3)+t_(4)))]`
C. `tan^(-1)[(g(t_(1)+t_(2)))/(a(t_(3)+t_(4)))]`
D. None of these

Explanation

To solve the problem, we need to analyze the motion of a projectile under the influence of gravity and horizontal retardation due to wind. We will derive the angle of projection based on the conditions given in the problem.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The projectile is launched with an initial velocity \( u \) at an angle \( \theta \) with respect to the horizontal.
- The horizontal component of the velocity is \( u \cos \theta \) and the vertical component is \( u \sin \theta \).
- Due to horizontal retardation \( a \), the horizontal motion will not be uniform.

2. **Vertical Motion Analysis**:
- The vertical position \( h \) of the projectile at time \( t \) is given by:
\[ h = u \sin \theta \cdot t - \frac{1}{2} g t^2 \]
- Since the projectile is at the same height at two different times \( t_1 \) and \( t_2 \), these times must satisfy the above equation. This implies that \( t_1 \) and \( t_2 \) are the roots of the quadratic equation formed.

3. **Forming the Quadratic Equation**:
- Rearranging the height equation gives:
\[ \frac{1}{2} g t^2 - u \sin \theta \cdot t + h = 0 \]
- The sum of the roots \( t_1 + t_2 \) can be found using the formula for the sum of roots of a quadratic equation:
\[ t_1 + t_2 = \frac{2u \sin \theta}{g} \]
- Let this be Equation (1).

4. **Horizontal Motion Analysis**:
- The horizontal distance \( x \) at time \( t \) is given by:
\[ x = u \cos \theta \cdot t - \frac{1}{2} a t^2 \]
- For the same horizontal distance at two different times \( t_3 \) and \( t_4 \), we can form a similar quadratic equation:
\[ \frac{1}{2} a t^2 - u \cos \theta \cdot t + x = 0 \]
- The sum of the roots \( t_3 + t_4 \) is:
\[ t_3 + t_4 = \frac{2u \cos \theta}{a} \]
- Let this be Equation (2).

5. **Finding the Angle of Projection**:
- We can now relate Equations (1) and (2):
\[ \frac{t_1 + t_2}{t_3 + t_4} = \frac{\frac{2u \sin \theta}{g}}{\frac{2u \cos \theta}{a}} \]
- Simplifying this gives:
\[ \frac{t_1 + t_2}{t_3 + t_4} = \frac{a \sin \theta}{g \cos \theta} \]
- Rearranging gives:
\[ \tan \theta = \frac{g}{a} \cdot \frac{t_1 + t_2}{t_3 + t_4} \]
- Thus, the angle of projection \( \theta \) can be expressed as:
\[ \theta = \tan^{-1} \left( \frac{g}{a} \cdot \frac{t_1 + t_2}{t_3 + t_4} \right) \]

### Final Result:
The angle of projection \( \theta \) is given by:
\[ \theta = \tan^{-1} \left( \frac{g}{a} \cdot \frac{t_1 + t_2}{t_3 + t_4} \right) \]