A cricket fielder can throw the cricket ball with a speed `v_(0)`.If the throws the ball while running with speed `u` at an angle `theta` to the horizontal.
The effective angle to the horizontal at which the ball is projected in air as seen by a spectator is

Options

A. `tan^(-1) [(v_(0)cos theta)/(u +v_(0) sin theta)]`
B. `tan^(-1) [(v_(0) sin theta)/(u+v_(0) cos theta)]`
C. `tan^(-)[(u)/(v_(0) cos theta + v_(0) sin theta)]`
D. `tan^(-1) [(v_(0) sin theta +v_(0) cos theta)/(u)]`

Explanation

To find the effective angle to the horizontal at which the ball is projected in the air as seen by a spectator, we can break down the problem step by step.

### Step-by-Step Solution:

1. **Identify the Components of the Velocity**:
- The fielder throws the ball with an initial speed \( v_0 \) at an angle \( \theta \) to the horizontal.
- The horizontal component of the throw is given by:
\[ v_{0x} = v_0 \cos \theta \]
- The vertical component of the throw is given by:
\[ v_{0y} = v_0 \sin \theta \]

2. **Consider the Fielder's Running Speed**:
- The fielder is running with a speed \( u \) horizontally.
- Therefore, the effective horizontal speed of the ball as seen by a spectator is the sum of the fielder's speed and the horizontal component of the throw:
\[ v_{effective\_x} = u + v_0 \cos \theta \]

3. **Determine the Effective Angle**:
- The effective angle \( \theta' \) that the ball makes with the horizontal can be determined using the tangent function:
\[ \tan \theta' = \frac{\text{Vertical Component}}{\text{Horizontal Component}} \]
- Substituting the components:
\[ \tan \theta' = \frac{v_0 \sin \theta}{u + v_0 \cos \theta} \]

4. **Calculate the Effective Angle**:
- To find \( \theta' \), we take the arctangent:
\[ \theta' = \tan^{-1} \left( \frac{v_0 \sin \theta}{u + v_0 \cos \theta} \right) \]

5. **Final Result**:
- The effective angle to the horizontal at which the ball is projected as seen by a spectator is:
\[ \theta' = \tan^{-1} \left( \frac{v_0 \sin \theta}{u + v_0 \cos \theta} \right) \]