In the following questions a statement of assertion (A) is followed by a statement of reason ( R).
A : In the case of ground to ground projection of a projectile from ground the angle of projection with horizontal is `theta =30^(@)` . There is no point on its path such that instantaneous velocity is normal to the initial velocity .
R : Maximum deviation of the projectile is `2theta = 60^(@)` .

Options

A. If both Assertion & Reason are true and the reason is the correct explanation of the assertion then mark (1) .
B. If both Assertion & Reason are true but the reason is not the correct explanation of the assertion then mark (2)
C. If Assertion is true statement but Reason is false then mark (3).
D. If both Assertion and Reason are false statements then mark (4) .

Answer

To solve the problem, we need to analyze both the assertion (A) and the reason (R) provided in the question regarding projectile motion.

### Step-by-Step Solution:

1. **Understanding the Assertion (A)**:
- The assertion states that for a projectile launched from the ground at an angle of \( \theta = 30^\circ \), there is no point in its trajectory where the instantaneous velocity is normal (perpendicular) to the initial velocity.
- In projectile motion, the initial velocity can be broken down into horizontal and vertical components. The initial velocity \( V_0 \) at \( 30^\circ \) can be expressed as:
\[ V_{0x} = V_0 \cos(30^\circ) \]
\[ V_{0y} = V_0 \sin(30^\circ) \]

2. **Analyzing the Trajectory**:
- As the projectile moves, its vertical component of velocity decreases until it reaches the maximum height and then increases again as it falls back down.
- At maximum height, the vertical velocity becomes zero, and the projectile only has horizontal velocity. The angle of the velocity vector changes throughout the flight but never becomes \( 60^\circ \) with respect to the initial velocity.

3. **Determining the Angle of Instantaneous Velocity**:
- The angle of the instantaneous velocity with respect to the horizontal changes continuously. At launch, it is \( 30^\circ \), and as it rises, it decreases until it reaches \( 0^\circ \) at the peak. After that, it increases back to \( 30^\circ \) when it hits the ground.
- Therefore, the angle of \( 60^\circ \) (which is \( 2 \times 30^\circ \)) is never achieved during the flight.

4. **Understanding the Reason (R)**:
- The reason states that the maximum deviation of the projectile is \( 2\theta = 60^\circ \).
- This is true in the context of the angle of projection. The maximum deviation refers to the angle between the initial velocity vector and the final velocity vector just before it hits the ground, which indeed is \( 60^\circ \).

5. **Conclusion**:
- Both the assertion (A) and the reason (R) are true. The reason correctly explains the assertion, as the maximum deviation of the projectile's angle is indeed \( 60^\circ \), but at no point does the instantaneous velocity become normal to the initial velocity.

### Final Answer:
- **Assertion (A)**: True
- **Reason (R)**: True
- The reason is a correct explanation for the assertion.