Two particles are projected with speeds `u_1` and `u_2` in mutually opposite directions from top of a tower horizontally. Find the time after which their velocity becomes perpendicular to each other.

Answer

To solve the problem of finding the time after which the velocities of two particles become perpendicular to each other, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Motion**:
- Two particles are projected horizontally with speeds \( u_1 \) and \( u_2 \) in opposite directions from the top of a tower.
- The horizontal component of their velocities remains constant since there is no horizontal acceleration.

2. **Vertical Motion**:
- Both particles are subject to gravitational acceleration. The vertical velocity \( v \) of each particle at any time \( t \) can be expressed as:
\[ v = -gt \]
- The negative sign indicates that the velocity is directed downward.

3. **Velocity Vectors**:
- For particle 1 (moving to the right with speed \( u_1 \)):
\[ \vec{v_1} = u_1 \hat{i} - gt \hat{j} \]
- For particle 2 (moving to the left with speed \( u_2 \)):
\[ \vec{v_2} = -u_2 \hat{i} - gt \hat{j} \]

4. **Condition for Perpendicularity**:
- The velocities of the two particles are perpendicular when their dot product is zero:
\[ \vec{v_1} \cdot \vec{v_2} = 0 \]

5. **Calculating the Dot Product**:
- Compute the dot product:
\[ (u_1 \hat{i} - gt \hat{j}) \cdot (-u_2 \hat{i} - gt \hat{j}) = u_1(-u_2) + (-gt)(-gt) \]
- This simplifies to:
\[ -u_1 u_2 + g^2 t^2 = 0 \]

6. **Solving for Time \( t \)**:
- Rearranging the equation gives:
\[ g^2 t^2 = u_1 u_2 \]
- Taking the square root of both sides:
\[ t = \frac{\sqrt{u_1 u_2}}{g} \]

### Final Answer:
The time after which their velocities become perpendicular to each other is:
\[ t = \frac{\sqrt{u_1 u_2}}{g} \]