I'm stuck on how to start this one problem. I am told the following: "A man walks at 5 km/h in the direction of a 20-km/h wind. Raindrops fall vertically at 6.6 km/h in still air. Determine the direction in which the drops appear to fall with respect to the man, measured clockwise from the direction of the wind." I know that for velocity comparisons, v_b = v_a + v_(b/a), but I don't know how to apply this logic for three different velocities in two different directions. Can someone please help me?

- anonymous

- schrodinger

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- anonymous

And before anyone else says it, I know this question would be better suited for the physics and engineering sections, but from my experience, nobody ever seems to check those for questions, so here I am.

- anonymous

Is anybody there?

- anonymous

Hello?

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## More answers

- anonymous

@Shalante , I notice you're reading my question. Could you perchance be able to help me out?

- anonymous

Nevermind, just saw your message. Will wait for further instruction.

- anonymous

Again, not looking for answers, here. Just want an idea of how to approach this.

- triciaal

|dw:1443676636690:dw|

- triciaal

vertical angles are equal
- tan theta = - tan ^-1 (6.6/20)

- triciaal

do you have answer choices?

- anonymous

No, just a set number of attempts.

- triciaal

do you have an idea what my sketch is?

- anonymous

From what I can tell, you've gor the speed of the man (6.6 km/h), and some arrow as (20 km/hr). Not sure where you got 20 from though. Also, tan^-1(6.6/20) is incorrect.

- anonymous

*got

- triciaal

|dw:1443677500413:dw|

- triciaal

you skipped the leading minus

- triciaal

it said measured clockwise

- anonymous

Shalante said they figured it out. Hold please,

- triciaal

@dan815 what is your take on this?

- anonymous

If two objects are traveling at the same direction, then the relative velocity is the difference between those two.
The wind would think the man is traveling 15km/h slower. The man would think the wind is traveling 15km/h faster.
With respect to is using the man as a reference.
|dw:1443677931462:dw|
So relative velocity of man 15km/h
Relative velocity of wind would be -15km/h
\[\tan^{-1} \frac{ opposite }{ adjacent }=\tan^{-1} \frac{ 6.6 }{ 15 }=\]
23.74 degrees

- anonymous

By the way, I am pretty active in the physics section, but not engineering though.

- anonymous

If two objects are traveling in opposite we would add them up.
Just like how cars move very fast in opposite direction.

- anonymous

Where'd you get 20 from though, when you were finding the difference? That's the only thing I don't get at this point. I get choosing the man as your reference point, but why'd you subtract 5 from 20? Where did 20 km/hr come from?

- anonymous

Wait, nevermind. I just realized my mistake.

- triciaal

@Shalante thanks I forgot to adjust for the man's velocity
@The_Silent_One the rain-drop (vertical) was 6.6
the wind was 20

- anonymous

I wrote the wrong value for that variable on my physical paper, so no wonder I've been getting this wrong.

- anonymous

No problem!
I got the angles messed up earlier.
I forgot to read that it was respect to man, so the angle is toward the x component.
I did this earlier
|dw:1443678723277:dw|
Thats why my first answer was wrong.

- anonymous

Okay, that makes sense. Thanks for helping me out! I gave you a medal for your explanation. I'm going to close this question now and get some sleep.

- anonymous

Yea, no problem.
Solving another difficult physics question right now.

- anonymous

Thanks for the medals

- anonymous

medal*

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