A boy is runing on the plane road with velocity v with a long hollow tube in his hand. The water is falling vertically downwards with velocity u. At what angle to the verticaly, he must inclined the tube the water drops enter it without touching its sides ?

Options

A. `tan^(-1)((v)/(u))`
B. `sin^(-1)((v)/(u))`
C. `tan^(-1)((u)/(v))`
D. `cos^(-1)((v)/(u))`

Explanation

To solve the problem of determining the angle at which the boy must incline the tube so that the water drops enter it without touching its sides, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Velocities**:
- The boy is running with a horizontal velocity \( v \) (to the right).
- The water is falling vertically downwards with a velocity \( u \).

2. **Define the Coordinate System**:
- Let the horizontal direction (to the right) be the positive x-direction.
- Let the vertical direction (downwards) be the negative y-direction.

3. **Express the Velocities in Vector Form**:
- The velocity of the boy (and the tube) can be represented as:
\[ \vec{v_p} = v \hat{i} \]
- The velocity of the water can be represented as:
\[ \vec{v_w} = -u \hat{j} \]

4. **Relative Velocity of Water with Respect to the Tube**:
- The velocity of the water with respect to the tube is given by:
\[ \vec{v_{wp}} = \vec{v_w} - \vec{v_p} = (-u \hat{j}) - (v \hat{i}) = -v \hat{i} - u \hat{j} \]

5. **Determine the Angle of Inclination**:
- The angle \( \theta \) that the tube makes with the vertical can be found using the relationship between the components of the relative velocity.
- The horizontal component of the relative velocity is \( -v \) (to the left) and the vertical component is \( -u \) (downwards).
- The angle \( \theta \) can be found using the tangent function:
\[ \tan \theta = \frac{\text{horizontal component}}{\text{vertical component}} = \frac{v}{u} \]
- Therefore, the angle \( \theta \) is given by:
\[ \theta = \tan^{-1}\left(\frac{v}{u}\right) \]

### Final Answer:
The angle \( \theta \) at which the tube must be inclined to the vertical is:
\[ \theta = \tan^{-1}\left(\frac{v}{u}\right) \]