Two balls are thrown horizontally from the top of a tower with velocities `v_(1)` and `v_(2)` in opposite directions at the same time. After how much time the angle between velocities of balls becomes `90^(@)` ?

Options

A. `(2sqrt(v_(1)v_(2)))/(g)`
B. `sqrt(v_(1)v_(2))/(g)`
C. `sqrt(v_(1)v_(2))/(2g)`
D. `(g)/sqrt(v_(1)v_(2))`

Explanation

To solve the problem of determining the time after which the angle between the velocities of two balls thrown horizontally from a tower becomes 90 degrees, we can follow these steps:

### Step-by-Step Solution

1. **Understanding the Motion**:
- Two balls are thrown horizontally from the top of a tower with velocities \( v_1 \) and \( v_2 \) in opposite directions.
- We can assume ball 1 is thrown in the positive x-direction and ball 2 in the negative x-direction.

2. **Velocity Components**:
- The velocity of ball 1 after time \( t \) can be expressed as:
\[ \vec{v_1} = v_1 \hat{i} - g t \hat{j} \]
- The velocity of ball 2 after time \( t \) can be expressed as:
\[ \vec{v_2} = -v_2 \hat{i} - g t \hat{j} \]

3. **Condition for Perpendicular Velocities**:
- The two velocities are perpendicular when their dot product is zero:
\[ \vec{v_1} \cdot \vec{v_2} = 0 \]
- Substituting the expressions for \( \vec{v_1} \) and \( \vec{v_2} \):
\[ (v_1 \hat{i} - g t \hat{j}) \cdot (-v_2 \hat{i} - g t \hat{j}) = 0 \]

4. **Calculating the Dot Product**:
- The dot product expands as follows:
\[ v_1(-v_2) + (-g t)(-g t) = 0 \]
- This simplifies to:
\[ -v_1 v_2 + g^2 t^2 = 0 \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ g^2 t^2 = v_1 v_2 \]

6. **Solving for Time \( t \)**:
- Taking the square root of both sides:
\[ t^2 = \frac{v_1 v_2}{g^2} \]
- Therefore, the time \( t \) when the velocities are perpendicular is:
\[ t = \sqrt{\frac{v_1 v_2}{g}} \]

### Final Answer
The time after which the angle between the velocities of the balls becomes \( 90^\circ \) is:
\[ t = \sqrt{\frac{v_1 v_2}{g}} \]