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 find the speed of the rain, we can break down the problem into steps based on the information given in the question.

### Step-by-Step Solution:

1. **Understanding the Situation**:
- The man moves down the inclined plane at a speed of \(5 \, \text{m/s}\) and sees the rain falling vertically.
- When moving up the incline at the same speed, he sees the rain making an angle \(\theta = \tan^{-1}(7/8)\) with the horizontal.

2. **Setting Up the Coordinate System**:
- Let the velocity of the rain be represented as \( \vec{v_r} = a \hat{i} + b \hat{j} \), where \(a\) is the horizontal component and \(b\) is the vertical component of the rain's velocity.

3. **Analyzing the First Case (Moving Down)**:
- When the man is moving down, his velocity can be broken into components:
\[ V_m = 5 \, \text{m/s} \text{ at } 37^\circ \]
- The horizontal component \(V_{mx} = 5 \cos(37^\circ) = 5 \cdot \frac{4}{5} = 4 \, \text{m/s}\)
- The vertical component \(V_{my} = 5 \sin(37^\circ) = 5 \cdot \frac{3}{5} = 3 \, \text{m/s}\)

- The man's velocity vector when moving down the incline is:
\[ \vec{V_m} = 4 \hat{i} + 3 \hat{j} \]

- Since the rain appears to fall vertically, the horizontal component of the rain's velocity relative to the man must be zero:
\[ a - 4 = 0 \implies a = 4 \]

4. **Analyzing the Second Case (Moving Up)**:
- When the man moves up the incline, his velocity vector is:
\[ \vec{V_m} = -4 \hat{i} - 3 \hat{j} \]

- The relative velocity of the rain with respect to the man is:
\[ \vec{V_{rm}} = \vec{V_r} - \vec{V_m} = (a + 4) \hat{i} + (b + 3) \hat{j} \]

- The angle \(\theta\) formed by the rain with the horizontal is given by:
\[ \tan(\theta) = \frac{\text{vertical component}}{\text{horizontal component}} = \frac{b + 3}{a + 4} \]
- Given \(\tan(\theta) = \frac{7}{8}\):
\[ \frac{b + 3}{a + 4} = \frac{7}{8} \]

5. **Substituting Known Values**:
- We already found \(a = 4\):
\[ \frac{b + 3}{4 + 4} = \frac{7}{8} \]
\[ \frac{b + 3}{8} = \frac{7}{8} \implies b + 3 = 7 \implies b = 4 \]

6. **Finding the Speed of the Rain**:
- Now we have \(a = 4\) and \(b = 4\). The speed of the rain can be calculated using the Pythagorean theorem:
\[ v_r = \sqrt{a^2 + b^2} = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \, \text{m/s} \]

### Final Answer:
The speed of the rain is \(4\sqrt{2} \, \text{m/s}\).