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 the same point with initial velocities \( u_1 \) and \( u_2 \) in opposite directions. The first particle moves to the right with velocity \( u_1 \), and the second particle moves to the left with velocity \( -u_2 \). Both particles are affected by gravity, which acts downwards.

### Step 2: Determine the velocity components
- For the first particle:
- Horizontal velocity: \( v_{1x} = u_1 \)
- Vertical velocity: \( v_{1y} = -gt \) (downward direction, hence negative)

- For the second particle:
- Horizontal velocity: \( v_{2x} = -u_2 \)
- Vertical velocity: \( v_{2y} = -gt \) (downward direction, hence negative)

### Step 3: Write the velocity vectors
The velocity vectors of the two particles can be expressed as:
- Particle 1: \( \vec{v_1} = u_1 \hat{i} - gt \hat{j} \)
- Particle 2: \( \vec{v_2} = -u_2 \hat{i} - gt \hat{j} \)

### Step 4: Use the condition for perpendicularity
For the two velocity vectors to be perpendicular, their dot product must equal zero:
\[ \vec{v_1} \cdot \vec{v_2} = 0 \]

### Step 5: Calculate the dot product
Calculating the dot product:
\[ \vec{v_1} \cdot \vec{v_2} = (u_1 \hat{i} - gt \hat{j}) \cdot (-u_2 \hat{i} - gt \hat{j}) \]
Expanding this:
\[ = u_1 \cdot (-u_2) + (-gt) \cdot (-gt) \]
\[ = -u_1 u_2 + g^2 t^2 \]

### Step 6: Set the dot product to zero
Setting the dot product to zero gives:
\[ -g_1 u_2 + g^2 t^2 = 0 \]
Rearranging this, we find:
\[ g^2 t^2 = u_1 u_2 \]

### Step 7: Solve for \( t \)
Taking the square root of both sides:
\[ t^2 = \frac{u_1 u_2}{g^2} \]
Thus,
\[ t = \sqrt{\frac{u_1 u_2}{g}} \]

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

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