Rain appears to be falling at `53^@` with the vertical to a man standing on the ground. When he starts running at 10 km/hr, rain again appears to be falling at `53^@` with the vertical. Find the speed (in km/hr) of rain wrt ground.

Options

A. 10
B. 5
C. `25/4`
D. `25/3`

Answer

To solve the problem, we need to analyze the situation step by step.

### Step 1: Understand the initial conditions
When the man is standing still, the rain appears to be falling at an angle of \(53^\circ\) with the vertical. This means that the velocity of the rain has a horizontal component and a vertical component.

### Step 2: Define the components of the rain's velocity
Let \(V_r\) be the speed of the rain with respect to the ground. The components of the rain's velocity can be expressed as:
- Vertical component: \(V_{r_y} = V_r \cos(53^\circ)\)
- Horizontal component: \(V_{r_x} = V_r \sin(53^\circ)\)

Using trigonometric values:
- \(\cos(53^\circ) = \frac{3}{5}\)
- \(\sin(53^\circ) = \frac{4}{5}\)

Thus, we can write:
- \(V_{r_y} = V_r \cdot \frac{3}{5}\)
- \(V_{r_x} = V_r \cdot \frac{4}{5}\)

### Step 3: Analyze the situation when the man starts running
When the man starts running at \(10 \text{ km/hr}\), the rain still appears to be falling at \(53^\circ\) with the vertical. Now, the horizontal component of the rain's velocity relative to the man becomes:
- \(V_{r_x} - V_m = V_r \cdot \frac{4}{5} - 10\)

### Step 4: Set up the relationship for the angle
Since the rain still appears to fall at \(53^\circ\) with the vertical, we can set up the tangent relationship:
\[ \tan(53^\circ) = \frac{V_{r_y}}{V_{r_x} - V_m} \]
Substituting the known values:
\[ \tan(53^\circ) = \frac{V_r \cdot \frac{3}{5}}{V_r \cdot \frac{4}{5} - 10} \]
Given that \(\tan(53^\circ) = \frac{4}{3}\), we can write:
\[ \frac{4}{3} = \frac{V_r \cdot \frac{3}{5}}{V_r \cdot \frac{4}{5} - 10} \]

### Step 5: Cross-multiply and simplify
Cross-multiplying gives:
\[ 4 \left(V_r \cdot \frac{4}{5} - 10\right) = 3 \left(V_r \cdot \frac{3}{5}\right) \]
Expanding both sides:
\[ \frac{16}{5} V_r - 40 = \frac{9}{5} V_r \]

### Step 6: Solve for \(V_r\)
Rearranging the equation:
\[ \frac{16}{5} V_r - \frac{9}{5} V_r = 40 \]
\[ \frac{7}{5} V_r = 40 \]
Multiplying both sides by \(\frac{5}{7}\):
\[ V_r = \frac{200}{7} \text{ km/hr} \approx 28.57 \text{ km/hr} \]

### Step 7: Final answer
Thus, the speed of the rain with respect to the ground is approximately \(28.57 \text{ km/hr}\).