Two particles are thrown horizontally in opposite directions with velocities u and 2u from the top of a high tower. The time after which their radius of curvature will be mutually perpendicular is

Options

A. `(sqrt2) u/g`
B. `2 u/g`
C. `1/(sqrt2) u/g`
D. `1/2 u/g`

Explanation

To solve the problem, we need to find the time after which the radii of curvature of two particles thrown horizontally from a tower will be mutually perpendicular. Let's break down the solution step by step.

### Step 1: Understanding the Motion
Two particles are thrown horizontally from the top of a tower with velocities \( u \) and \( 2u \) in opposite directions. The motion can be analyzed in two dimensions: horizontal (x-direction) and vertical (y-direction).

### Step 2: Velocity Components
For the first particle (velocity \( u \)):
- Horizontal velocity: \( v_{1x} = u \)
- Vertical velocity after time \( t \): \( v_{1y} = gt \) (downward)

Thus, the velocity vector for the first particle is:
\[ \mathbf{v_1} = u \hat{i} - gt \hat{j} \]

For the second particle (velocity \( 2u \)):
- Horizontal velocity: \( v_{2x} = -2u \) (negative because it's in the opposite direction)
- Vertical velocity after time \( t \): \( v_{2y} = gt \) (downward)

Thus, the velocity vector for the second particle is:
\[ \mathbf{v_2} = -2u \hat{i} - gt \hat{j} \]

### Step 3: Condition for Perpendicular Velocities
The velocities \( \mathbf{v_1} \) and \( \mathbf{v_2} \) are perpendicular when their dot product is zero:
\[ \mathbf{v_1} \cdot \mathbf{v_2} = 0 \]

Calculating the dot product:
\[ \mathbf{v_1} \cdot \mathbf{v_2} = (u)(-2u) + (-gt)(-gt) = -2u^2 + g^2t^2 \]

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

### Step 4: Solve for Time \( t \)
Rearranging the equation gives:
\[ g^2t^2 = 2u^2 \]
\[ t^2 = \frac{2u^2}{g^2} \]
Taking the square root of both sides:
\[ t = \frac{u}{g} \sqrt{2} \]

### Conclusion
The time after which the radius of curvature of the two particles will be mutually perpendicular is:
\[ t = \frac{u}{g} \sqrt{2} \]