A man moves on a cycle with a velocity of 4 km/hr. The rain appears to fall on him with a velocity of 3 km/hr vertically. The actual velocity of the rain is

Options

A. 7 km/hr
B. 5 km/hr
C. 4/3 km/hr
D. 3/4 km/hr

Answer

To find the actual velocity of the rain, we can use vector addition. Here's a step-by-step solution:

### Step 1: Understand the problem
The man is cycling with a velocity of 4 km/hr horizontally (let's assume to the right) and perceives the rain falling vertically with a velocity of 3 km/hr. We need to find the actual velocity of the rain.

### Step 2: Set up the vectors
- The velocity of the man (Vm) can be represented as:
\[ \vec{V_m} = 4 \hat{i} \text{ km/hr} \]
- The apparent velocity of the rain with respect to the man (Vr_m) is:
\[ \vec{V_{r/m}} = -3 \hat{j} \text{ km/hr} \]
(the negative sign indicates that it is falling downwards).

### Step 3: Use vector addition
The actual velocity of the rain (Vr) can be found using the equation:
\[ \vec{V_r} = \vec{V_m} + \vec{V_{r/m}} \]
Substituting the vectors:
\[ \vec{V_r} = 4 \hat{i} - 3 \hat{j} \text{ km/hr} \]

### Step 4: Calculate the magnitude of the actual velocity of the rain
To find the magnitude of the actual velocity of the rain, we can use the Pythagorean theorem:
\[ |\vec{V_r}| = \sqrt{(V_{r,x})^2 + (V_{r,y})^2} \]
Where \(V_{r,x} = 4 \text{ km/hr}\) and \(V_{r,y} = -3 \text{ km/hr}\):
\[ |\vec{V_r}| = \sqrt{(4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ km/hr} \]

### Step 5: Calculate the angle of the rain's velocity
To find the angle θ that the velocity vector makes with the horizontal, we can use the tangent function:
\[ \tan(\theta) = \frac{V_{r,y}}{V_{r,x}} = \frac{-3}{4} \]
Thus, the angle θ can be found using:
\[ \theta = \tan^{-1}\left(\frac{-3}{4}\right) \]
Calculating this gives:
\[ \theta \approx -36.87^\circ \]
This angle indicates that the rain is falling at an angle of approximately 36.87 degrees below the horizontal.

### Final Answer
The actual velocity of the rain is 5 km/hr at an angle of approximately 36.87 degrees below the horizontal.

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