Two particles move in a uniform gravitational field with an acceleration g. At the initial moment the particles were located over a tower at one point and moved with velocities `v_1 = 3m//s and v_2= 4m//s` horizontally in opposite directions. Find the distance between the particles at the moment when their velocity vectors become mutually perpendicular.

Explanation

To solve the problem of finding the distance between two particles moving in a uniform gravitational field when their velocity vectors become mutually perpendicular, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Initial Conditions:**
- Two particles are thrown horizontally from the same height with initial velocities:
- Particle 1: \( v_1 = 3 \, \text{m/s} \) (to the right)
- Particle 2: \( v_2 = 4 \, \text{m/s} \) (to the left)

2. **Velocity Components:**
- The horizontal component of the velocities remains constant since there is no horizontal acceleration:
- Particle 1: \( \vec{V_1} = 3 \hat{i} - gt \hat{j} \)
- Particle 2: \( \vec{V_2} = -4 \hat{i} - gt \hat{j} \)

3. **Condition for Perpendicular Vectors:**
- The velocity vectors are perpendicular when their dot product is zero:
\[ \vec{V_1} \cdot \vec{V_2} = 0 \]
- This gives:
\[ (3 \hat{i} - gt \hat{j}) \cdot (-4 \hat{i} - gt \hat{j}) = 0 \]
- Expanding the dot product:
\[ 3 \cdot (-4) + (-gt) \cdot (-gt) = 0 \]
\[ -12 + g^2 t^2 = 0 \]
- Rearranging gives:
\[ g^2 t^2 = 12 \]
\[ t^2 = \frac{12}{g^2} \]
\[ t = \frac{\sqrt{12}}{g} \]

4. **Substituting the Value of g:**
- Given \( g = 10 \, \text{m/s}^2 \):
\[ t = \frac{\sqrt{12}}{10} = \frac{2\sqrt{3}}{10} = \frac{\sqrt{3}}{5} \, \text{s} \]

5. **Calculating Horizontal Displacements:**
- Horizontal displacement of Particle 1:
\[ x_1 = v_1 \cdot t = 3 \cdot \frac{\sqrt{3}}{5} = \frac{3\sqrt{3}}{5} \, \text{m} \]
- Horizontal displacement of Particle 2:
\[ x_2 = v_2 \cdot t = 4 \cdot \frac{\sqrt{3}}{5} = \frac{4\sqrt{3}}{5} \, \text{m} \]

6. **Finding the Distance Between the Particles:**
- The total distance between the two particles when they are perpendicular:
\[ \text{Distance} = x_1 + x_2 = \frac{3\sqrt{3}}{5} + \frac{4\sqrt{3}}{5} = \frac{7\sqrt{3}}{5} \, \text{m} \]

### Final Answer:
The distance between the two particles when their velocity vectors become mutually perpendicular is:
\[ \frac{7\sqrt{3}}{5} \, \text{m} \]