(A): When a body is projected at angle `45^(@)` its range is maximum
(R): For maximum of range, the value of `sin 2 theta` should be equal to one.

Options

A. Both (A) and (R) are true and (R) is the correct explanation of (A)
B. Both (A) and (R) are true and (R) is not the correct explanation of (A)
C. (A) is true but (R) is false
D. Both (A) and (R) are fals

Answer

To solve the question, we need to analyze the statements provided and understand the physics behind projectile motion.

### Step-by-Step Solution:

1. **Understanding the Assertion (A)**:
- The assertion states that when a body is projected at an angle of \(45^\circ\), its range is maximum.
- In projectile motion, the range \(R\) is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
where \(u\) is the initial velocity, \(\theta\) is the angle of projection, and \(g\) is the acceleration due to gravity.

2. **Analyzing the Range Formula**:
- From the range formula, we see that the range depends on \(\sin 2\theta\).
- To maximize the range, we need to maximize \(\sin 2\theta\).

3. **Finding the Maximum Value of \(\sin 2\theta\)**:
- The maximum value of \(\sin x\) is 1. Therefore, for maximum range:
\[ \sin 2\theta = 1 \]
- This occurs when:
\[ 2\theta = 90^\circ \quad \text{or} \quad \theta = 45^\circ \]

4. **Conclusion on Assertion**:
- Since we have shown that projecting a body at \(45^\circ\) gives the maximum range, the assertion (A) is correct.

5. **Understanding the Reason (R)**:
- The reason states that for maximum range, the value of \(\sin 2\theta\) should be equal to one.
- We have already established that this is true as it leads to the angle of projection being \(45^\circ\).

6. **Conclusion on Reason**:
- The reason (R) is also correct.

7. **Final Conclusion**:
- Both the assertion (A) and the reason (R) are correct. Thus, the answer is that both A and R are correct.

### Final Answer:
Both assertion (A) and reason (R) are correct.

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