Assertion A particle is projected with speed u at an angle `theta` with the horizontal. At any time during motion, speed of particle is v at angle `alpha` with the vertical, then `v sin alpha` is always constant throughout the motion.
Reason In case of projectile motion, magnitude of radical acceleration at topmost point is maximum.

Options

A. If both Asseration and Reason are correct and Reason is the correct explanation of Assertion
B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion
C. If Assertion is true but Reason is false
D. If Assertion is false but Reason is true

Explanation

To solve the question, we need to analyze both the assertion and the reason provided.

### Step-by-Step Solution:

1. **Understanding the Assertion**:
- A particle is projected with an initial speed \( u \) at an angle \( \theta \) with the horizontal.
- During its motion, at any time, the speed of the particle is \( v \) at an angle \( \alpha \) with the vertical.
- We need to show that \( v \sin \alpha \) is constant throughout the motion.

2. **Analyzing the Components of Velocity**:
- The velocity \( v \) can be broken down into two components:
- The horizontal component: \( v \cos \alpha \)
- The vertical component: \( v \sin \alpha \)
- In projectile motion, the horizontal component of velocity remains constant because there is no horizontal acceleration (ignoring air resistance).

3. **Horizontal Motion**:
- The horizontal component of the initial velocity is \( u \cos \theta \).
- Since there is no horizontal acceleration, this component remains constant throughout the motion.
- Thus, \( v \cos \alpha = u \cos \theta \) is constant.

4. **Vertical Motion**:
- The vertical component of velocity changes due to gravitational acceleration. However, the horizontal component's constancy leads us to focus on the relationship between \( v \) and \( \alpha \).

5. **Using the Relationship**:
- Since \( v \cos \alpha \) is constant, we can express \( v \) in terms of \( \alpha \):
\[ v = \frac{u \cos \theta}{\cos \alpha} \]
- Now, we can find \( v \sin \alpha \):
\[ v \sin \alpha = \frac{u \cos \theta}{\cos \alpha} \sin \alpha = u \cos \theta \tan \alpha \]
- As \( \tan \alpha \) varies with the angle of the particle, \( v \sin \alpha \) remains constant because it is derived from the constant horizontal component.

6. **Conclusion for Assertion**:
- Therefore, \( v \sin \alpha \) is indeed constant throughout the motion, confirming that the assertion is correct.

7. **Understanding the Reason**:
- The reason states that at the topmost point of projectile motion, the magnitude of radial acceleration is maximum.
- Radial acceleration is given by the formula:
\[ a_r = g^2 - a_t^2 \]
- At the topmost point, the tangential acceleration \( a_t \) is zero, thus:
\[ a_r = g \]
- This means that the radial acceleration is indeed maximum at the topmost point.

8. **Conclusion for Reason**:
- While both the assertion and reason are correct, the reason does not correctly explain the assertion about \( v \sin \alpha \) being constant.

### Final Answer:
Both the assertion and the reason are correct, but the reason is not the correct explanation of the assertion.