A person can swim at `4 km *h^(-1)` in still water . At what angle should he set himself to cross the river in a direction perpendicular to the river current of velocity `2 km *h^(-1)`.

Explanation

Let the velocity of the person be `vecv` velocity of the current be `vecu` and the resultant of `vecu and vecv be vecw` [Fig.2.51]. The person crosses the river along its width from one bank to the other. Hence, `vecw and vecu` are perpendicular to each other. If the angle between `vecv and vecw is theta` , then from Fig.2.51,
`sin theta =u/v=2/4=1/2 or, theta =30^@`.
Hence, the person has to swin at an inclination of `30^@` with the width of the river, or at an angle of `(90^@+30^@) or 120^@` with the direction of the current as shown in Fig.2.51.