Two particles are simultaneously projected in opposite directions horizontally from a given point in space where gravity g is uniform. If `u_(1)` and `u_(2)` be their initial speeds, then the time t after which their velocitites are mutually perpendicular is given by

Options

A. `sqrt(u_(1)u_(2))/(g)`
B. `sqrt(u_(1)^(2)+u_(2)^(2))/(g)`
C. `sqrt(u_(1)(u_(1)+u_(2)))/(g)`
D. `sqrt(u_(2)(u_(1)+u_(2)))/(g)`

Explanation

To solve the problem of finding the time \( t \) after which the velocities of two particles projected in opposite directions are mutually perpendicular, we can follow these steps:

### Step 1: Understand the Motion of the Particles
Two particles are projected horizontally from a point with initial speeds \( u_1 \) and \( u_2 \) in opposite directions. Due to gravity, both particles will experience vertical acceleration downwards.

### Step 2: Determine the Velocity Components
For each particle, we can break down the velocity into horizontal and vertical components:

- **Particle 1** (projected with speed \( u_1 \)):
- Horizontal component: \( v_{1x} = -u_1 \) (to the left)
- Vertical component: \( v_{1y} = g t \) (downwards, starting from rest)

- **Particle 2** (projected with speed \( u_2 \)):
- Horizontal component: \( v_{2x} = u_2 \) (to the right)
- Vertical component: \( v_{2y} = g t \) (downwards, starting from rest)

### Step 3: Write the Velocity Vectors
The velocity vectors for the two particles can be expressed as:
- \( \vec{v_1} = -u_1 \hat{i} + g t \hat{j} \)
- \( \vec{v_2} = u_2 \hat{i} + g t \hat{j} \)

### Step 4: Condition for Perpendicular Velocities
The velocities are mutually perpendicular when their dot product is zero:
\[ \vec{v_1} \cdot \vec{v_2} = 0 \]

Calculating the dot product:
\[ (-u_1 \hat{i} + g t \hat{j}) \cdot (u_2 \hat{i} + g t \hat{j}) = -u_1 u_2 + (g t)(g t) = 0 \]

### Step 5: Simplify the Equation
From the dot product equation, we have:
\[ -u_1 u_2 + g^2 t^2 = 0 \]
Rearranging gives:
\[ g^2 t^2 = u_1 u_2 \]

### Step 6: Solve for Time \( t \)
Taking the square root of both sides, we find:
\[ t^2 = \frac{u_1 u_2}{g^2} \]
Thus,
\[ t = \sqrt{\frac{u_1 u_2}{g}} \]

### Conclusion
The time \( t \) after which the velocities of the two particles are mutually perpendicular is given by:
\[ t = \sqrt{\frac{u_1 u_2}{g}} \]