Two particles move in a uniform gravitational field with an acceleration `g`.At the initial moment the particles were located at same point and moved with velocities `u_(1)=9 ms^(-1)` and `u_(2)=4 ms^(-1)` horizontally in opposite directions.The time between the particles at the moment when their velocity vectors are mutually perpendicular in `s` in (take `g=10 ms^(-2)`)

Options

A. `0.36`
B. `3.6`
C. `0.6`
D. 6

Explanation

To solve the problem, we need to determine the time at which the velocity vectors of two particles moving in a gravitational field become mutually perpendicular. Let's break down the solution step by step.

### Step 1: Understand the Initial Conditions
Both particles start at the same point and move horizontally in opposite directions with initial velocities:
- Particle 1: \( u_1 = 9 \, \text{m/s} \) (to the right)
- Particle 2: \( u_2 = 4 \, \text{m/s} \) (to the left)

### Step 2: Determine the Velocity Components
Since both particles are moving under the influence of gravity, their vertical velocity components will change over time while their horizontal components remain constant.

For Particle 1:
- Horizontal velocity: \( v_{1x} = 9 \, \text{m/s} \)
- Vertical velocity: \( v_{1y} = 0 - g \cdot t = -10t \, \text{m/s} \) (downward)

For Particle 2:
- Horizontal velocity: \( v_{2x} = -4 \, \text{m/s} \) (since it moves in the opposite direction)
- Vertical velocity: \( v_{2y} = 0 - g \cdot t = -10t \, \text{m/s} \) (downward)

### Step 3: Set Up the Condition for Perpendicular Vectors
The velocity vectors of the two particles are mutually perpendicular when their dot product is zero. The dot product of two vectors \( \mathbf{A} = (A_x, A_y) \) and \( \mathbf{B} = (B_x, B_y) \) is given by:
\[ \mathbf{A} \cdot \mathbf{B} = A_x B_x + A_y B_y \]

For our particles:
\[ v_1 \cdot v_2 = (9)(-4) + (-10t)(-10t) = 0 \]
This simplifies to:
\[ -36 + 100t^2 = 0 \]

### Step 4: Solve for Time \( t \)
Rearranging the equation gives us:
\[ 100t^2 = 36 \]
\[ t^2 = \frac{36}{100} = 0.36 \]
Taking the square root:
\[ t = \sqrt{0.36} = 0.6 \, \text{s} \]

### Step 5: Conclusion
The time at which the velocity vectors of the two particles are mutually perpendicular is \( t = 0.6 \, \text{s} \).

### Summary
- The two particles start at the same point and move horizontally in opposite directions.
- Their vertical velocities change due to gravity.
- The condition for their velocity vectors to be perpendicular leads us to find that \( t = 0.6 \, \text{s} \).