When a man moves down the inclined plane with a constant speed `5 ms^-1` which makes an angle of `37^@` with the horizontal, he finds that the rain is falling vertically downward. When he moves up the same inclined plane with the same speed, he finds that the rain makes an angle `theta = tan^-1 (7/8)` with the horizontal. The speed of the rain is

Options

A. `sqrt(116) ms^-1`
B. `sqrt(32) ms^-1`
C. `5 ms^-1`
D. `sqrt(73) ms^-1`

Explanation

To solve the problem, we need to analyze the motion of the man and the rain in two different scenarios: when the man is moving down the inclined plane and when he is moving up the inclined plane.

### Step-by-Step Solution:

1. **Understanding the Scenario**:
- The man moves down the inclined plane at a speed of \(5 \, \text{m/s}\) at an angle of \(37^\circ\) with the horizontal.
- When moving down, he observes that the rain is falling vertically downward.
- When moving up, he observes that the rain makes an angle \(\theta = \tan^{-1}\left(\frac{7}{8}\right)\) with the horizontal.

2. **Components of Velocity**:
- The components of the man's velocity when moving down the incline:
- \(V_{mx} = 5 \cos(37^\circ) = 5 \times \frac{4}{5} = 4 \, \text{m/s}\) (horizontal component)
- \(V_{my} = 5 \sin(37^\circ) = 5 \times \frac{3}{5} = 3 \, \text{m/s}\) (vertical component)

3. **Rain Falling Vertically**:
- When the man moves down and sees the rain falling vertically, the horizontal component of the rain's velocity must equal the negative of the man's horizontal velocity:
\[ V_{rx} + 4 = 0 \implies V_{rx} = -4 \, \text{m/s} \]
- The vertical component of the rain's velocity is \(V_{ry}\) (unknown).

4. **Moving Up the Incline**:
- When the man moves up the incline, his velocity components are:
- \(V_{mx} = 4 \, \text{m/s}\) (same as before)
- \(V_{my} = -3 \, \text{m/s}\) (upward direction)
- The relative velocity of the rain with respect to the man is:
\[ V_{rx} - 4 \quad \text{and} \quad V_{ry} + 3 \]
- The angle of the rain with respect to the horizontal is given by:
\[ \tan(\theta) = \frac{V_{ry} + 3}{V_{rx} - 4} = \frac{7}{8} \]

5. **Substituting Known Values**:
- We already have \(V_{rx} = -4\):
\[ \tan(\theta) = \frac{V_{ry} + 3}{-4 - 4} = \frac{V_{ry} + 3}{-8} \]
- Setting this equal to \(\frac{7}{8}\):
\[ \frac{V_{ry} + 3}{-8} = \frac{7}{8} \]
- Cross-multiplying gives:
\[ V_{ry} + 3 = -7 \implies V_{ry} = -10 \, \text{m/s} \]

6. **Calculating the Speed of the Rain**:
- The speed of the rain can be calculated using the Pythagorean theorem:
\[ \text{Speed of rain} = \sqrt{V_{rx}^2 + V_{ry}^2} = \sqrt{(-4)^2 + (-10)^2} = \sqrt{16 + 100} = \sqrt{116} = \sqrt{4 \times 29} = 2\sqrt{29} \, \text{m/s} \]

### Final Answer:
The speed of the rain is \(2\sqrt{29} \, \text{m/s}\).