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 find the relationship between the ranges \( R_1 \) and \( R_2 \) of two projectiles thrown at angles of \( 30^\circ \) and \( 60^\circ \) respectively, both with the same initial velocity.

### Step 1: Understanding the Angles
The first projectile is thrown at an angle of \( 30^\circ \) with the horizontal, while the second projectile is thrown at an angle of \( 30^\circ \) with the vertical. This means that the second projectile is thrown at an angle of \( 60^\circ \) with the horizontal (since \( 90^\circ - 30^\circ = 60^\circ \)).

### Step 2: Formula for Range of a Projectile
The formula for the range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \( u \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

### Step 3: Calculate \( R_1 \)
For the first projectile thrown at \( 30^\circ \):
\[ R_1 = \frac{u^2 \sin(2 \times 30^\circ)}{g} = \frac{u^2 \sin(60^\circ)}{g} = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} = \frac{u^2 \sqrt{3}}{2g} \]

### Step 4: Calculate \( R_2 \)
For the second projectile thrown at \( 60^\circ \):
\[ R_2 = \frac{u^2 \sin(2 \times 60^\circ)}{g} = \frac{u^2 \sin(120^\circ)}{g} = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} = \frac{u^2 \sqrt{3}}{2g} \]

### Step 5: Compare \( R_1 \) and \( R_2 \)
From the calculations, we see that:
\[ R_1 = R_2 \]
Thus, the relationship between the ranges \( R_1 \) and \( R_2 \) is:
\[ R_1 = R_2 \]

### Step 6: Conclusion
Both projectiles, despite being thrown at different angles, have the same range when thrown with the same initial velocity.