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 1: Define the Variables
Let:
- \( u_A \) = initial velocity of ball A (projected vertically upwards)
- \( u_B \) = initial velocity of ball B (projected at \( 30^\circ \) to the vertical)

### Step 2: Determine the Time of Flight for Ball A
For ball A, which is projected vertically upwards:
- The time taken to reach the maximum height \( t \) can be calculated using the equation of motion:
\[ v = u_A - g t \]
At maximum height, the final velocity \( v = 0 \):
\[ 0 = u_A - g t \implies t = \frac{u_A}{g} \]
- The total time of flight \( T_A \) for ball A (up and down) is:
\[ T_A = 2t = 2 \cdot \frac{u_A}{g} = \frac{2u_A}{g} \]

### Step 3: Determine the Time of Flight for Ball B
For ball B, which is projected at \( 30^\circ \) to the vertical:
- The angle with the horizontal is \( 60^\circ \) (since \( 90^\circ - 30^\circ = 60^\circ \)).
- The time of flight \( T_B \) for projectile motion is given by:
\[ T_B = \frac{2u_B \sin \theta}{g} = \frac{2u_B \sin 60^\circ}{g} \]
Using \( \sin 60^\circ = \frac{\sqrt{3}}{2} \):
\[ T_B = \frac{2u_B \cdot \frac{\sqrt{3}}{2}}{g} = \frac{\sqrt{3}u_B}{g} \]

### Step 4: Set the Times of Flight Equal
Since both balls reach the ground simultaneously, we set \( T_A = T_B \):
\[ \frac{2u_A}{g} = \frac{\sqrt{3}u_B}{g} \]
Cancelling \( g \) from both sides:
\[ 2u_A = \sqrt{3}u_B \]

### Step 5: Solve for the Ratio of Velocities
Rearranging the equation gives us:
\[ \frac{u_A}{u_B} = \frac{\sqrt{3}}{2} \]

### Conclusion
The ratio of the velocities of projection of balls A and B is:
\[ u_A : u_B = \sqrt{3} : 2 \]