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November 3rd, 2019, 07:16 PM  #1 
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 178 Thanks: 5 Math Focus: Calculus  How can I find the speed and the angle of a water droplet?
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 $345$ 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. Last edited by skipjack; November 3rd, 2019 at 07:49 PM. 
November 3rd, 2019, 07:53 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 21,106 Thanks: 2324 
Vector subtraction gives the relative velocity of the droplets.

November 4th, 2019, 03:37 AM  #3  
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 178 Thanks: 5 Math Focus: Calculus  Quote:
I'm assuming that $\vec{u}=\textrm{runner vector}$ and $\vec{v}=\textrm{water droplet vector}$ $\left \ \vec{v}\vec{u} \right \=\sqrt{4.5^2+6^2}=\frac{15}{2}=7.5$ The angle I presume can be obtained from taking the inverse tangent function: $\omega=\tan^{1}\left(\frac{4.5}{6}\right)=37^{\circ}$ And these correspond to that of the first alternative and it checks with the given answer. But again why?. In this particular situation I'm assuming that the modulus of the difference between these vectors is the same as the addition. But when I attempted to think about this problem I thought that the rain droplets as seen by the runner would "bend" and form some sort of diagonal of the right triangle hence I used $6$ the given speed for the hypotenuse of that triangle. But it doesn't seem to be the case. I know it mentions vertical, but would this "assumed" bend effect doesn't happen as seen by the runner?. Can you help me to clarify this doubt please?.  
November 4th, 2019, 05:12 AM  #4 
Senior Member Joined: Jun 2019 From: USA Posts: 376 Thanks: 202 
Let $\hat{i}$ be horizontal and $\hat{j}$ be vertical. Imagine you're standing there watching the runner and the rain. From your point of view, the runner is moving at $4.5~m/s ~\hat{i}$ and the rain is moving at $6~m/s ~\hat{j}$. Now imagine things from the runner's point of view. You (and the earth) appear to be moving at $4.5~m/s ~\hat{i}$. The rain is simultaneously moving backwards and falling, with a net velocity (from the runner's point of view) of $4.5~m/s ~\hat{i} 6~m/s ~\hat{j}$. 

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angle, droplet, find, speed, water 
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