Two projectiles are fired from the same point with the same speed at angles of projection `60^(@) (B) and 30^(@) (A)` respectively. Which one of the following is true?

Options

A. `R_(A) = R_(B)`
B. `H_(B) = 3H_(A)`
C. `T_(B) = sqrt(3)T_(A)`
D. None of these

Explanation

To solve the problem, we will analyze the motion of two projectiles fired at different angles but with the same initial speed. We will calculate the time of flight, horizontal range, and maximum height for both projectiles.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Initial speed (u) is the same for both projectiles.
- Angles of projection:
- Projectile A (θ_A) = 30°
- Projectile B (θ_B) = 60°

2. **Calculate the Time of Flight (T):**
The time of flight (T) for a projectile is given by the formula:
\[ T = \frac{2u \sin \theta}{g} \]

- For Projectile A:
\[ T_A = \frac{2u \sin(30°)}{g} = \frac{2u \cdot \frac{1}{2}}{g} = \frac{u}{g} \]

- For Projectile B:
\[ T_B = \frac{2u \sin(60°)}{g} = \frac{2u \cdot \frac{\sqrt{3}}{2}}{g} = \frac{\sqrt{3}u}{g} \]

3. **Compare the Times of Flight:**
- From the calculations:
\[ T_B = \sqrt{3} T_A \]
This means that the time of flight for Projectile B is greater than that of Projectile A.

4. **Calculate the Horizontal Range (R):**
The horizontal range (R) is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]

- For Projectile A:
\[ R_A = \frac{u^2 \sin(60°)}{g} = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} = \frac{\sqrt{3}u^2}{2g} \]

- For Projectile B:
\[ R_B = \frac{u^2 \sin(120°)}{g} = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} = \frac{\sqrt{3}u^2}{2g} \]

5. **Compare the Ranges:**
- From the calculations:
\[ R_A = R_B \]
This means that both projectiles have the same horizontal range.

6. **Calculate the Maximum Height (H):**
The maximum height (H) is given by:
\[ H = \frac{u^2 \sin^2(\theta)}{2g} \]

- For Projectile A:
\[ H_A = \frac{u^2 \sin^2(30°)}{2g} = \frac{u^2 \cdot \left(\frac{1}{2}\right)^2}{2g} = \frac{u^2}{8g} \]

- For Projectile B:
\[ H_B = \frac{u^2 \sin^2(60°)}{2g} = \frac{u^2 \cdot \left(\frac{\sqrt{3}}{2}\right)^2}{2g} = \frac{3u^2}{8g} \]

7. **Compare the Maximum Heights:**
- From the calculations:
\[ H_B = 3 H_A \]
This means that the maximum height attained by Projectile B is three times that of Projectile A.

### Conclusion:
- **Range:** \( R_A = R_B \)
- **Time of Flight:** \( T_B = \sqrt{3} T_A \)
- **Height:** \( H_B = 3 H_A \)

Thus, the correct options are:
1. \( R_A = R_B \)
2. \( H_B = 3 H_A \)
3. \( T_B = \sqrt{3} T_A \)