Assertion : Two projectile have maximum heights 4H and H respectively. The ratio of their horizontal components of velocities should be 1:2 for their horizontal ranges to be same.
Reason : Horizontal range = horizontal component of velocity xx time of flight.

Options

A. (a)If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.
B. (b)If both Assertion and Reason are true and the Reason is not the correct explanation of the Assertion.
C. (c) If Assertion is true, but the Reason is false.
D. (d) If Assertion is false, but the Reason is true.

Explanation

To solve the given problem, we need to analyze the assertion and reason provided regarding the projectile motion of two projectiles with different maximum heights.

### Step-by-Step Solution:

1. **Understanding Maximum Heights**:
- Let the maximum height of the first projectile be \( H_1 = 4H \).
- Let the maximum height of the second projectile be \( H_2 = H \).

2. **Using the Formula for Maximum Height**:
- The formula for the maximum height \( H \) of a projectile is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
- For the first projectile:
\[ 4H = \frac{u_1^2 \sin^2 \theta_1}{2g} \]
- For the second projectile:
\[ H = \frac{u_2^2 \sin^2 \theta_2}{2g} \]

3. **Relating the Velocities**:
- From the equations above, we can express \( u_1 \sin \theta_1 \) and \( u_2 \sin \theta_2 \):
\[ u_1 \sin \theta_1 = \sqrt{8gH} \]
\[ u_2 \sin \theta_2 = \sqrt{2gH} \]

4. **Calculating Time of Flight**:
- The time of flight \( T \) for a projectile is given by:
\[ T = \frac{2u \sin \theta}{g} \]
- For the first projectile:
\[ T_1 = \frac{2u_1 \sin \theta_1}{g} = \frac{2 \sqrt{8gH}}{g} = \frac{4\sqrt{2H}}{g} \]
- For the second projectile:
\[ T_2 = \frac{2u_2 \sin \theta_2}{g} = \frac{2 \sqrt{2gH}}{g} = \frac{2\sqrt{2H}}{g} \]

5. **Calculating Horizontal Range**:
- The horizontal range \( R \) is given by:
\[ R = u_x \cdot T \]
- For the first projectile:
\[ R_1 = u_{x1} \cdot T_1 \]
- For the second projectile:
\[ R_2 = u_{x2} \cdot T_2 \]

6. **Setting Ranges Equal**:
- For the ranges to be equal \( R_1 = R_2 \):
\[ u_{x1} \cdot T_1 = u_{x2} \cdot T_2 \]
- Rearranging gives:
\[ \frac{u_{x1}}{u_{x2}} = \frac{T_2}{T_1} \]

7. **Substituting Time Values**:
- Substitute \( T_1 \) and \( T_2 \):
\[ \frac{u_{x1}}{u_{x2}} = \frac{\frac{2\sqrt{2H}}{g}}{\frac{4\sqrt{2H}}{g}} = \frac{1}{2} \]

8. **Conclusion**:
- The ratio of the horizontal components of velocities \( u_{x1} : u_{x2} = 1 : 2 \).
- Therefore, the assertion is true, and the reason is also true as it correctly states the relationship between horizontal range, horizontal component of velocity, and time of flight.

### Final Answer:
Both the assertion and the reason are true.