Two particles move in a uniform gravitational field with an acceleration g. At the intial moment the particles were located at one point and moved with velocity `v_(1)=1" "ms^(-1)" and "v_(2)=4" "ms^(-1)` horizontally in opposite directions. Find the time interval after their velocity vectors become mutually perpendicular

Answer

To solve the problem, we need to analyze the motion of the two particles under the influence of gravity and determine when their velocity vectors become mutually perpendicular. Here’s a step-by-step solution:

### Step 1: Define the initial velocities
Let’s denote the velocities of the two particles:
- Particle 1 has an initial velocity \( \vec{v_1} = -1 \hat{i} \) m/s (moving in the negative x-direction).
- Particle 2 has an initial velocity \( \vec{v_2} = 4 \hat{i} \) m/s (moving in the positive x-direction).

### Step 2: Determine the effect of gravity on the velocities
Since both particles are moving in a uniform gravitational field, the vertical component of their velocities will change over time due to gravity. The vertical component of velocity for both particles after time \( t \) will be:
- For Particle 1: \( v_{1y} = -gt \)
- For Particle 2: \( v_{2y} = -gt \)

Thus, the velocity vectors after time \( t \) become:
- \( \vec{v_1}(t) = -1 \hat{i} - gt \hat{j} \)
- \( \vec{v_2}(t) = 4 \hat{i} - gt \hat{j} \)

### Step 3: Set up the condition for perpendicularity
For the two velocity vectors to be mutually perpendicular, their dot product must be zero:
\[ \vec{v_1}(t) \cdot \vec{v_2}(t) = 0 \]

Calculating the dot product:
\[ (-1 \hat{i} - gt \hat{j}) \cdot (4 \hat{i} - gt \hat{j}) = (-1)(4) + (-gt)(-gt) = -4 + g^2 t^2 \]

Setting the dot product equal to zero:
\[ -4 + g^2 t^2 = 0 \]

### Step 4: Solve for time \( t \)
Rearranging the equation gives:
\[ g^2 t^2 = 4 \]
\[ t^2 = \frac{4}{g^2} \]
\[ t = \frac{2}{g} \]

### Step 5: Substitute the value of \( g \)
Assuming \( g = 10 \, \text{m/s}^2 \):
\[ t = \frac{2}{10} = 0.2 \, \text{s} \]

### Conclusion
The time interval after which the velocity vectors of the two particles become mutually perpendicular is \( t = 0.2 \, \text{s} \).

---