A projectile thrown at an angle of `30^@` with the horizontal has a range `R_1` and attains a maximum height `h_1` Another projectile thrown, with the same velocity, at an angle `30^@` with the vertical, has a range `R_2` and attains a maximum height `h_2` The relation between `R_1 and R_2 ` is

Options

A. `R_1 =(R_2)/2`
B. `R_1 =R_2`
C. `R_1 = 2R_2`
D. `R_1 =4 R_2`

Explanation

To solve the problem, we need to analyze the motion of two projectiles thrown at different angles but with the same initial velocity.

### Step 1: Analyze the first projectile
The first projectile is thrown at an angle of \(30^\circ\) with the horizontal.

- **Initial Velocity Components**:
- Horizontal component: \( u_x = u \cos(30^\circ) = u \cdot \frac{\sqrt{3}}{2} \)
- Vertical component: \( u_y = u \sin(30^\circ) = u \cdot \frac{1}{2} \)

- **Time of Flight**:
The time of flight \( T_1 \) can be calculated using the formula:
\[ T_1 = \frac{2u_y}{g} = \frac{2 \cdot \left( u \cdot \frac{1}{2} \right)}{g} = \frac{u}{g} \]

- **Range**:
The range \( R_1 \) is given by:
\[ R_1 = u_x \cdot T_1 = \left( u \cdot \frac{\sqrt{3}}{2} \right) \cdot \left( \frac{u}{g} \right) = \frac{u^2 \sqrt{3}}{2g} \]

- **Maximum Height**:
The maximum height \( h_1 \) can be calculated using:
\[ h_1 = \frac{u_y^2}{2g} = \frac{\left( u \cdot \frac{1}{2} \right)^2}{2g} = \frac{u^2}{8g} \]

### Step 2: Analyze the second projectile
The second projectile is thrown at an angle of \(30^\circ\) with the vertical, which means it is thrown at an angle of \(60^\circ\) with the horizontal.

- **Initial Velocity Components**:
- Horizontal component: \( u_x = u \cos(60^\circ) = u \cdot \frac{1}{2} \)
- Vertical component: \( u_y = u \sin(60^\circ) = u \cdot \frac{\sqrt{3}}{2} \)

- **Time of Flight**:
The time of flight \( T_2 \) is:
\[ T_2 = \frac{2u_y}{g} = \frac{2 \cdot \left( u \cdot \frac{\sqrt{3}}{2} \right)}{g} = \frac{u \sqrt{3}}{g} \]

- **Range**:
The range \( R_2 \) is given by:
\[ R_2 = u_x \cdot T_2 = \left( u \cdot \frac{1}{2} \right) \cdot \left( \frac{u \sqrt{3}}{g} \right) = \frac{u^2 \sqrt{3}}{2g} \]

- **Maximum Height**:
The maximum height \( h_2 \) can be calculated using:
\[ h_2 = \frac{u_y^2}{2g} = \frac{\left( u \cdot \frac{\sqrt{3}}{2} \right)^2}{2g} = \frac{3u^2}{8g} \]

### Step 3: Compare \( R_1 \) and \( R_2 \)
From the calculations, we see that:
\[ R_1 = \frac{u^2 \sqrt{3}}{2g} \]
\[ R_2 = \frac{u^2 \sqrt{3}}{2g} \]

Thus, we find that:
\[ R_1 = R_2 \]

### Conclusion
The relation between the ranges \( R_1 \) and \( R_2 \) is:
\[ R_1 = R_2 \]