A man wishes to swim across a river 40 m wide flowing with a speed of 3m/s. such that he reaches the point just infront on the other bank in time not greater than 10s. The angle made by the direction he swims and river flow direction is :-

Options

A. `90^(@)`
B. `127^(@)`
C. `150^(@)`
D. `143^(@)`

Answer

To solve the problem, we need to determine the angle made by the direction the man swims and the river flow direction. Let's break down the solution step by step.

### Step 1: Understand the problem
The river is 40 meters wide and flows with a speed of 3 m/s. The swimmer wants to reach directly across the river in a time not greater than 10 seconds.

### Step 2: Identify the components of motion
Let:
- \( v \) = speed of the swimmer
- \( \theta \) = angle made by the swimmer's direction with the river flow direction
- \( v_x = v \cos \theta \) = horizontal component of the swimmer's speed
- \( v_y = v \sin \theta \) = vertical component of the swimmer's speed

### Step 3: Set up the equations
1. The swimmer must reach the opposite bank in 10 seconds. The width of the river is 40 m, so:
\[ v_y \cdot t = 40 \quad \text{(1)} \]
Substituting \( t = 10 \) seconds:
\[ v \sin \theta \cdot 10 = 40 \implies v \sin \theta = 4 \quad \text{(2)} \]

2. To reach directly across without drifting downstream, the horizontal component of the swimmer's speed must equal the speed of the river:
\[ v_x = v \cos \theta = 3 \quad \text{(3)} \]

### Step 4: Solve the equations
Now we have two equations:
1. \( v \sin \theta = 4 \)
2. \( v \cos \theta = 3 \)

Dividing equation (2) by equation (3):
\[ \frac{v \sin \theta}{v \cos \theta} = \frac{4}{3} \]
This simplifies to:
\[ \tan \theta = \frac{4}{3} \]

### Step 5: Calculate the angle
To find \( \theta \):
\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \]
Calculating this gives:
\[ \theta \approx 53^\circ \]

### Step 6: Determine the angle with respect to the river flow
Since the angle made by the direction he swims and the river flow direction is \( \theta \), we need to find the angle \( \alpha \) that is complementary to \( \theta \):
\[ \text{Angle with respect to river flow} = 180^\circ - \theta \]
Thus:
\[ \text{Angle} = 180^\circ - 53^\circ = 127^\circ \]

### Final Answer
The angle made by the direction he swims and the river flow direction is \( 127^\circ \).

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