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`

Answer

To solve the problem, we will use the principles of projectile motion. We need to find the speed \( v \) of a particle thrown with an initial speed \( u \) at an angle \( \theta \) with the horizontal when it makes an angle \( \phi \) with the horizontal.

### Step-by-Step Solution:

1. **Identify the Components of Initial Velocity**:
- The initial velocity \( u \) can be resolved into two components:
- Horizontal component: \( u_x = u \cos \theta \)
- Vertical component: \( u_y = u \sin \theta \)

2. **Consider the Components of Final Velocity**:
- When the particle makes an angle \( \phi \) with the horizontal, its velocity \( v \) can also be resolved into components:
- Horizontal component: \( v_x = v \cos \phi \)
- Vertical component: \( v_y = v \sin \phi \)

3. **Conservation of Horizontal Velocity**:
- In projectile motion, the horizontal component of velocity remains constant (assuming no air resistance). Therefore:
\[ u \cos \theta = v \cos \phi \]

4. **Rearranging the Equation**:
- From the equation above, we can express \( v \) in terms of \( u \), \( \theta \), and \( \phi \):
\[ v = \frac{u \cos \theta}{\cos \phi} \]

5. **Final Expression**:
- The final speed \( v \) of the particle when it makes an angle \( \phi \) with the horizontal is given by:
\[ v = u \frac{\cos \theta}{\cos \phi} \]