A particle is thrown with a speed u at an angle ` theta` with the horizontal. When the particle makes an angle `phi` with the horizontal. Its speed changes to v :

Options

A. `v = u cos theta`
B. `v = u cos theta cos phi`
C. `v = u cos theta sec phi`
D. `v = u sec theta cos phi`

Explanation

To solve the problem, we need to establish the relationship between the initial speed \( u \) and the final speed \( v \) of a particle thrown at an angle \( \theta \) with the horizontal, when it makes an angle \( \phi \) with the horizontal.

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- The particle is thrown with an initial speed \( u \) at an angle \( \theta \) with the horizontal.

2. **Determine the Horizontal Component of the Initial Velocity**:
- The horizontal component of the initial velocity \( u \) is given by:
\[ u_x = u \cos \theta \]

3. **Determine the Horizontal Component of the Final Velocity**:
- When the particle makes an angle \( \phi \) with the horizontal, its speed changes to \( v \). The horizontal component of the final velocity \( v \) is:
\[ v_x = v \cos \phi \]

4. **Apply the Principle of Conservation of Horizontal Velocity**:
- Since there are no horizontal forces acting on the particle, the horizontal component of the velocity remains constant. Therefore, we can equate the two horizontal components:
\[ u \cos \theta = v \cos \phi \]

5. **Rearranging the Equation**:
- From the equation above, we can solve for \( v \):
\[ v = \frac{u \cos \theta}{\cos \phi} \]
- This can also be expressed using the secant function:
\[ v = u \cos \theta \cdot \sec \phi \]

6. **Final Result**:
- The relationship between the final speed \( v \) and the initial speed \( u \) is:
\[ v = u \cos \theta \sec \phi \]

### Conclusion:
The final expression shows how the speed of the particle changes based on the angle of projection and the angle at which it is observed.