# How can I find the speed and the angle of a water droplet in a rainy day as seen by a person running?

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.

• The $3-4-5$ triangle is a hint as to the nature of the result. He is traveling horizontally, the droplets travel vertically, the resulting relative speed must combine these quantities in some manner. Since the two lines of travel are perpendicular, it makes sense to think about the hypotenuse between them... Nov 4 '19 at 3:43
• @abiessu Perhaps can you develop an answer so I could understand what you mean by combining the speeds?. Can you show me the equation should I use?. Had I not been given the alternatives, what should I have done with this question? Nov 4 '19 at 4:27

• water droplet speed as seen from a stationary point on earth is $\vect{v_d}=(0,-6)$. the speed of the runner as seen from the same point is $\vect{v_r}=(4.5,0)$. however, from the perspective of the runner, the water droplet both falls down with the speed $\vect{v_d}=(0,-6)$ and it moves towards him with the horizontal speed $\vect{v_h}=(-4.5,0)$. wherefore, for the runner the water moves with a resultant speed $\vect{v_r}=\vect{v_d}+\vect{v_h}=(0,-6)+(-4.5,0)=(-4.5,-6)$. this vector has a magnidue of 7.5 and a direction of $37^0$ West from vertical. Nov 4 '19 at 6:49