A swimmer wishes to reach directly opposite bank of a river, flowing with velocity 8 m/s. The swimmer can swim 10 m/s still water. The width of the river is 480 m. Time taken by him to do so:

Options

A. 60 sec
B. 48 sec
C. 80 sec
D. None of these.

Answer

To solve the problem of the swimmer trying to reach the opposite bank of a river flowing with a velocity of 8 m/s, while he can swim at a speed of 10 m/s in still water, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Velocity of the river, \( V_r = 8 \, \text{m/s} \)
- Velocity of the swimmer in still water, \( V_s = 10 \, \text{m/s} \)
- Width of the river, \( d = 480 \, \text{m} \)

2. **Determine the Angle of Swimming:**
- To reach directly opposite the bank, the swimmer must swim at an angle \( \theta \) such that the resultant velocity has no x-component (i.e., it should be directed straight across the river).
- The x-component of the swimmer's velocity must balance the river's velocity:
\[ V_r = V_s \cos \theta \]
- This gives us:
\[ 8 = 10 \cos \theta \]
- Rearranging gives:
\[ \cos \theta = \frac{8}{10} = 0.8 \]
- Therefore, \( \theta = \cos^{-1}(0.8) \).

3. **Calculate the y-component of the Swimmer's Velocity:**
- The y-component of the swimmer's velocity is given by:
\[ V_{sy} = V_s \sin \theta \]
- Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - 0.8^2} = \sqrt{1 - 0.64} = \sqrt{0.36} = 0.6 \]
- Thus:
\[ V_{sy} = 10 \sin \theta = 10 \times 0.6 = 6 \, \text{m/s} \]

4. **Calculate the Time Taken to Cross the River:**
- The time taken to cross the river can be calculated using the formula:
\[ \text{Time} = \frac{\text{Distance}}{\text{Velocity}} = \frac{d}{V_{sy}} = \frac{480 \, \text{m}}{6 \, \text{m/s}} = 80 \, \text{s} \]

### Final Answer:
The time taken by the swimmer to reach the opposite bank is **80 seconds**.

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