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 `prop` with the bank. The flag on the mast of the launch makes an angle `beta` with its direction of motion. Determine the speed of the launch w.r.t. the bank. The velocity of wind is `u` perpendicular to the stream.
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Answer 1

Correct Answer - `(u sin(prop + beta - pi//2))/(sin beta)`.
Flag on a launch 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.
`vec V_("Wind,Earth") = vec u = u hat j`
`vec v_("Launch,Earth") = vec v`
=`v cos prop hat i+ v sin prop hat j`
`vec V_("Wind, Launch") = vec u - vec v`
=`-v cos prop hat i+ (u - v sin prop) hat j`
`tan(prop + beta) = (u - v sin prop)/(-v cos prop)`
`-v tan(prop + beta) cos prop = u - v sin prop`
`v[sin prop - cos prop tan(prop + beta)] = u`
`v=(-u cos(prop + beta))/(sin beta) = (u sin(prop + beta - pi//2))/(sin beta)`.
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