Rankers Physics
Topic: Kinematics
Subtopic: Relative Motion in Two Dimension

A man wishes to swim across a river 0.5 km wide. If he can swim at the rate of 2 km/h in still water and the river flows at the rate of 1 km/h. The angle made by the directon (w.r.t. the flow of the river) along which he should swim so as to reach a point exactly opposite his starting point, should be :
60°
120°
145°
90°

Solution:

To swim across the river and reach a point directly opposite his starting point, the man must swim at an angle against the flow of the river to counteract the river's current.

### Given:
- Speed of the man in still water = 2 km/h
- Speed of the river flow = 1 km/h

Let \(\theta\) be the angle between the man's swimming direction and the river flow. To counteract the river's flow, the horizontal component of the man's velocity must equal the river's velocity.

\[
\text{Horizontal component} = 2 \sin \theta = 1
\]

Solving for \(\theta\):

\[
\sin \theta = \frac{1}{2}
\]
\[
\theta = \sin^{-1} \left( \frac{1}{2} \right) = 30^\circ
\]

Thus, the man should swim at an angle of:

\[
{120^\circ} \text{ upstream from the river flow.}
\]

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