(B) Let $V_{sw} = 12 \, km/h$ be the velocity of the swimmer in still water and $v_r = 6 \, km/h$ be the velocity of the river flow.To reach the point

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(B) Let $V_{sw} = 12 \, km/h$ be the velocity of the swimmer in still water and $v_r = 6 \, km/h$ be the velocity of the river flow.To reach the point directly opposite to the starting point,the swimmer must swim at an angle $\theta$ with the normal to the river flow such that the component of his velocity cancels the river flow velocity.$V_{sw} \sin \theta = v_r$$12 \sin \theta = 6$$\sin \theta = \frac{6}{12} = \frac{1}{2}$$\theta = 30^{\circ}$ (angle with the normal).The angle $\alpha$ with respect to the direction of the river flow is given by $\alpha = 90^{\circ} + \theta = 90^{\circ} + 30^{\circ} = 120^{\circ}$.

Class 11 Physics 3-2.Motion in Plane Relative Velocity (river boat, rain, wind)

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$A$ swimmer can swim with a velocity of $12 \, km/h$ in still water. The water flowing in a river has a velocity of $6 \, km/h$. The angle with respect to the direction of flow of the river water at which the swimmer should swim in order to reach the point on the other bank just opposite to his starting point is ........$^{\circ}$. (Round off to the Nearest Integer)

JEE MAIN 2021 All JEE MAIN

A
$30$
B
$120$
C
$60$
D
$150$

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3-2.Motion in Plane

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Relative Velocity (river boat, rain, wind)

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$A$ boat takes $2 \, h$ to travel $8 \, km$ and back in still water. If the velocity of water is $4 \, km/h$,the time taken for going upstream $8 \, km$ and coming back is:

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