Two particles are projected from a tower horizontally in opposite directions with velocities `10 m//s and 20 m//s`. Find the time when their velocity vectors are mutually perpendicular. Take `g=10m//s^2`.

Explanation

To solve the problem of finding the time when the velocity vectors of two particles projected horizontally from a tower become mutually perpendicular, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Initial Velocities**:
- Let the first particle be projected with an initial velocity \( u_1 = 20 \, \text{m/s} \) in the positive x-direction.
- Let the second particle be projected with an initial velocity \( u_2 = -10 \, \text{m/s} \) in the negative x-direction.

2. **Determine the Velocity Components**:
- For the first particle:
- Horizontal component: \( v_{1x} = u_1 = 20 \, \text{m/s} \)
- Vertical component: \( v_{1y} = -gt = -10t \, \text{m/s} \) (since \( g = 10 \, \text{m/s}^2 \))
- Therefore, the velocity vector of the first particle after time \( t \) is:
\[ \mathbf{v_1} = 20 \hat{i} - 10t \hat{j} \]

- For the second particle:
- Horizontal component: \( v_{2x} = u_2 = -10 \, \text{m/s} \)
- Vertical component: \( v_{2y} = -gt = -10t \, \text{m/s} \)
- Therefore, the velocity vector of the second particle after time \( t \) is:
\[ \mathbf{v_2} = -10 \hat{i} - 10t \hat{j} \]

3. **Condition for Perpendicular Vectors**:
- The two velocity vectors are mutually perpendicular when their dot product is zero:
\[ \mathbf{v_1} \cdot \mathbf{v_2} = 0 \]

4. **Calculate the Dot Product**:
- Calculate the dot product:
\[ \mathbf{v_1} \cdot \mathbf{v_2} = (20 \hat{i} - 10t \hat{j}) \cdot (-10 \hat{i} - 10t \hat{j}) \]
- Expanding the dot product:
\[ = 20 \cdot (-10) + (-10t) \cdot (-10t) \]
\[ = -200 + 100t^2 \]

5. **Set the Dot Product to Zero**:
- Set the equation to zero:
\[ -200 + 100t^2 = 0 \]
- Rearranging gives:
\[ 100t^2 = 200 \]
\[ t^2 = 2 \]
\[ t = \sqrt{2} \, \text{s} \]

6. **Conclusion**:
- The time when the velocity vectors are mutually perpendicular is:
\[ t = \sqrt{2} \, \text{s} \approx 1.41 \, \text{s} \]