A river is flowing at the rate of 6 km/hr. A swimmer swims across with a velocity of 9 km/hr w.r.t. water. The resultant velocity of the man will be in (km/hr):-

Options

A. `sqrt(117)`
B. `sqrt(340)`
C. `sqrt(17)`
D. `3sqrt(40)`

Answer

To solve the problem of finding the resultant velocity of a swimmer swimming across a river, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the velocities**:
- The velocity of the river (\(V_r\)) is 6 km/hr.
- The velocity of the swimmer with respect to the water (\(V_s\)) is 9 km/hr.

2. **Set up the coordinate system**:
- Assume the river flows along the x-axis (horizontal direction).
- The swimmer swims across the river, which we can assume to be along the y-axis (vertical direction).

3. **Represent the velocities as vectors**:
- The velocity of the swimmer can be represented as a vector: \(V_s = 9 \hat{j}\) km/hr (where \(\hat{j}\) is the unit vector in the y-direction).
- The velocity of the river can be represented as a vector: \(V_r = 6 \hat{i}\) km/hr (where \(\hat{i}\) is the unit vector in the x-direction).

4. **Calculate the resultant velocity**:
- The resultant velocity (\(V_{resultant}\)) of the swimmer can be found by adding the two vectors:
\[ V_{resultant} = V_s + V_r = 9 \hat{j} + 6 \hat{i} \]

5. **Find the magnitude of the resultant velocity**:
- To find the magnitude of the resultant velocity, we use the Pythagorean theorem:
\[ |V_{resultant}| = \sqrt{(V_r)^2 + (V_s)^2} = \sqrt{(6)^2 + (9)^2} \]
- Calculate the squares:
\[ = \sqrt{36 + 81} = \sqrt{117} \]

6. **Calculate the numerical value**:
- The numerical value of \(\sqrt{117}\) can be approximated:
\[ \sqrt{117} \approx 10.82 \text{ km/hr} \]

### Final Answer:
The resultant velocity of the swimmer is approximately **10.82 km/hr**.

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