Two balls A and B are projected from the same location simultaneously. Ball A is projected vertically upwards and ball B at ` 30^@`
to the vertical. They reach the ground simultaneously. The velocities of projection of A and B are in the ratio

Options

A. `sqrt3:1`
B. `1: sqrt3`
C. `sqrt3:2`
D. `2: sqrt3`

Explanation

To solve the problem of finding the ratio of the velocities of projection of balls A and B, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Parameters**:
- Let \( u_A \) be the velocity of projection of ball A (projected vertically upwards).
- Let \( u_B \) be the velocity of projection of ball B (projected at an angle of \( 30^\circ \) to the vertical).

2. **Time of Flight for Ball A**:
- The time of flight \( T_A \) for ball A, which is projected vertically upwards, is given by the formula:
\[ T_A = \frac{2u_A \sin(90^\circ)}{g} = \frac{2u_A}{g} \]
- Here, \( \sin(90^\circ) = 1 \).

3. **Time of Flight for Ball B**:
- The time of flight \( T_B \) for ball B, which is projected at an angle of \( 30^\circ \) to the vertical, is given by:
\[ T_B = \frac{2u_B \sin(30^\circ)}{g} \]
- Here, \( \sin(30^\circ) = \frac{1}{2} \), thus:
\[ T_B = \frac{2u_B \cdot \frac{1}{2}}{g} = \frac{u_B}{g} \]

4. **Equating the Time of Flights**:
- Since both balls reach the ground simultaneously, we have:
\[ T_A = T_B \]
\[ \frac{2u_A}{g} = \frac{u_B}{g} \]

5. **Canceling \( g \)**:
- We can cancel \( g \) from both sides:
\[ 2u_A = u_B \]

6. **Finding the Ratio**:
- Rearranging gives us:
\[ \frac{u_A}{u_B} = \frac{1}{2} \]

7. **Final Ratio**:
- Therefore, the ratio of the velocities of projection of A and B is:
\[ u_A : u_B = 1 : 2 \]

### Conclusion:
The velocities of projection of A and B are in the ratio \( 1:2 \).

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