If a body 'A' of mass M is thrown with velocity V at an angle of `30^(@)` to the horizontal and another body B of the same mass is thrown with the same speed at an angle of `60^(@)` to the horizontal, the ratio of horizontal ranges of A to B will be

Options

A. `1 : 3`
B. `1 : 1`
C. `1 : sqrt(3)`
D. `sqrt(3) : 1`

Explanation

To solve the problem of finding the ratio of horizontal ranges of two bodies A and B thrown at different angles, we can follow these steps:

### Step 1: Understand the Range Formula
The horizontal range \( R \) of a projectile 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,
- \( g \) is the acceleration due to gravity.

### Step 2: Identify the Angles and Initial Velocity
In this problem:
- Body A is thrown at an angle of \( 30^\circ \) with velocity \( V \).
- Body B is thrown at an angle of \( 60^\circ \) with the same velocity \( V \).

### Step 3: Calculate the Range for Both Bodies
For body A:
\[ R_A = \frac{V^2 \sin(2 \times 30^\circ)}{g} = \frac{V^2 \sin(60^\circ)}{g} \]

For body B:
\[ R_B = \frac{V^2 \sin(2 \times 60^\circ)}{g} = \frac{V^2 \sin(120^\circ)}{g} \]

### Step 4: Find the Sine Values
We know:
- \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \)
- \( \sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2} \)

### Step 5: Set Up the Ratio of Ranges
Now, we can set up the ratio of the ranges:
\[ \frac{R_A}{R_B} = \frac{\frac{V^2 \sin(60^\circ)}{g}}{\frac{V^2 \sin(120^\circ)}{g}} = \frac{\sin(60^\circ)}{\sin(120^\circ)} \]

### Step 6: Simplify the Ratio
Since \( \sin(60^\circ) = \sin(120^\circ) \):
\[ \frac{R_A}{R_B} = \frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}} = 1 \]

### Conclusion
The ratio of the horizontal ranges of A to B is:
\[ \frac{R_A}{R_B} = 1 \]

### Final Answer
Thus, the ratio of horizontal ranges of A to B is \( 1:1 \).

---