# application of vectors: parachute in the wind

Parachute in the wind. In still air, a parachute with a payload would fall vertically at terminal speed of 4 m/s. Find the direction and magnitude of its terminal velocity relative to the ground if it falls in a stead wind blowing horizontally west to east at 10 m/s.

So I went about this problem by drawing a right triangle. I had 10 m/s be the horizontal component and 4 m/s be the vertical component. I applied the pythagorean theorem and found the magnitude of its terminal velocity relative to the ground to be 10.8 m/s.

Then to find the direction I did intense tangent for the horizontal and vertical components and got 21.8 degrees. I am just now very confused how I find the direction like is it North of east, South of east..etc. Can someone please explain step by step?

We are adding two vectors which are perpendicular to each other; one with magnitude $4$ and direction downwards and the other with magnitude $10$ and direction eastward.
You are correct in drawing the right triangle and the magnitude calculation, but perhaps it would help to understand where the triangle is drawn to find the direction. Let us examine the right triangle with the eastward vector and a translation of the downward vector east as legs. Using basic trigonometry, we find the angle is $\arctan\left(\frac{4}{10}\right),$ which is about $21.8$ degrees. Looking at the triangle, we see that this angle is below east, since the parachute is losing altitude.
Our final answer is $\boxed{21.8^{\circ}}$ below east at a speed of $\boxed{10.8\frac{\text{m}}{\text{s}}}.$