The problem is as follows:
A person is running following a constant speed of $4.5\,\frac{m}{s}$ over a flat (horizontal) track on a rainy day. The water droplets fall vertically with a measured speed of $6\,\frac{m}{s}$. Find the speed in $\frac{m}{s}$ of the water droplet as seen by the person running. Find the angle to the vertical should his umbrella be inclined to get wet the less possible. (You may use the relationship of $37^{\circ}-53^{\circ}-90^{\circ}$ for the $3-4-5$ right triangle ).
The alternatives given on my book are as follows:
$\begin{array}{ll} 1.7.5\,\frac{m}{s};\,37^{\circ}\\ 2.7.5\,\frac{m}{s};\,53^{\circ}\\ 3.10\,\frac{m}{s};\,37^{\circ}\\ 4.10\,\frac{m}{s};\,53^{\circ}\\ 5.12.5\,\frac{m}{s};\,37^{\circ}\\ \end{array}$
On this problem I'm really very lost. What sort of equation should I use to get the vectors or the angles and most importantly to get the relative speed which is what is being asked.
I assume that to find the relative speed can be obtaining by subtracting the speed from which the water droplet is falling to what the person is running. But He is in this case running horizontally. How can I subtract these?
Can somebody give me some help here?
Supposedly the answer is the first option or $1$. But I have no idea how to get there.
Since the rain is vertical, the runner runs into falling droplets. To him is like the droplets fall slanted towards him.
You have a right triangle formed with the opposite of the speed of the runner and the speed of the droplet as legs abs the resulting speed of the droplet as the hypothenuse.
The first answer is correct.