Question

A launch travels across a river from a point A to a point B of the opposite bank along the line AB forming angle $\alpha$ with the bank. The flag on the mast of the launch makes an angle $\beta$ with its a direction of motion. Determine the speed of the launch w.r.t. the bank. The velocity of wind is $u$ perpendicular to the stream.

Answer 1

Flag on a launch flies in the direction of the wind relative to the launch. Therefore, $\beta$ is the angle made by relative velocity of the wind in relation to the launch.

${\overline{v}}_{wind,Earth}=\overline{u}=u\overline{j}$

${\overline{v}}_{Launch,Earth}=\overline{v}=v\cos \alpha \overline{i}+v\sin \alpha \hat{j}$

${\overline{V}}_{Wind,Launch}=\overline{u}-\overline{v}=-v\cos \alpha \hat{i}+(u-v\sin \alpha )\hat{j}$

$\tan (\alpha +\beta )=\frac{u-vsin\alpha }{-v\cos \alpha }$

$-v\tan (\alpha +\beta )=\frac{u-v\sin \alpha }{-v\cos \alpha }$

$v[\sin \alpha --\cos \alpha \tan (\alpha +\beta \left)\right]=u$

$v=\frac{-u\cos (\alpha +\beta )}{\sin \beta }=\frac{u\sin (\alpha +\beta -\pi /2)}{\sin \beta }$