Hello, missashley!

A ship cruises forward at $\displaystyle V_s = 5$ m/s relative to the water.

On deck, a man walks diagonally toward the bow such that his path forms

an angle $\displaystyle \theta = 22^o$ with a line perpendicular to the boat's direction of motion.

He walks at $\displaystyle V_m = 2$ m/s relative to the boat.

The speed he walks relative to the water is 5.749 m/s.

At what angle to his intended path does the man walk with respect to the water?

Answer in degrees. Code:

B C
* - - → *
↑ /
| /
| / 2
|22*/
| /
| /
|/
*
A

In the right triangle, we have: .$\displaystyle \begin{array}{ccccc}AB & = & 2\cos22^o & \approx & 1.854 \\

BC & = & 2\sin22^o & \approx & 0.749\end{array}$

Relative to the ship, he is walking 0.749 m/sec forward.

Since the ship is moving forward at 5 m/sec,

. . the man is moving forward at 5.749 m/sec (relative to the water).

(That's where they got that number.)

So relative to the water, his movement looks like this: Code:

5.749
* - - - - - *
| /
| /
1.854 | /
| /
|θ/
*

We have: .$\displaystyle \tan\theta \:=\:\frac{5.749}{1.854} \:=\:3.100862999$

Hence: .$\displaystyle \theta\:=\:\tan^{-1}(3.100863) \:\approx\:72.1^o$

Relative to the forward direction of the boat, his angle is about 17.9°.