A particle is thrown with a speed is at an angle `theta` with the horizontal. When the particle makes an angle `phi` with the horizontal, its speed changes to v, then

Options

A. `v = v_(0) cos theta`
B. `v = v_(0) cos theta *cos phi`
C. `v = v_(0) cos theta * sec phi`
D. `v = v_(0) sec theta * cos phi`

Explanation

To solve the problem step by step, we need to analyze the motion of the particle thrown at an angle and how its velocity changes when it reaches a different angle.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- The particle is thrown with an initial speed \( V_0 \) at an angle \( \theta \) with the horizontal.

2. **Determine Horizontal Velocity**:
- The horizontal component of the initial velocity can be calculated using trigonometry:
\[ V_{0x} = V_0 \cos \theta \]

3. **Identify Conditions at Angle \( \phi \)**:
- When the particle makes an angle \( \phi \) with the horizontal, its speed changes to \( V \).
- The horizontal component of this new velocity is:
\[ V_x = V \cos \phi \]

4. **Apply the Principle of Horizontal Velocity Conservation**:
- Since there is no horizontal acceleration (assuming no air resistance), the horizontal component of velocity remains constant:
\[ V_{0x} = V_x \]
- Therefore, we can equate the two expressions for horizontal velocity:
\[ V_0 \cos \theta = V \cos \phi \]

5. **Solve for \( V \)**:
- Rearranging the equation gives us:
\[ V = \frac{V_0 \cos \theta}{\cos \phi} \]
- This can also be expressed using the secant function:
\[ V = V_0 \cos \theta \sec \phi \]

6. **Final Result**:
- Thus, the final expression for the speed \( V \) of the particle when it makes an angle \( \phi \) with the horizontal is:
\[ V = V_0 \cos \theta \sec \phi \]