anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • 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
  • anonymous
Is anybody there?
anonymous
  • anonymous
Hello?

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anonymous
  • anonymous
@Shalante , I notice you're reading my question. Could you perchance be able to help me out?
anonymous
  • anonymous
Nevermind, just saw your message. Will wait for further instruction.
anonymous
  • anonymous
Again, not looking for answers, here. Just want an idea of how to approach this.
triciaal
  • triciaal
|dw:1443676636690:dw|
triciaal
  • triciaal
vertical angles are equal - tan theta = - tan ^-1 (6.6/20)
triciaal
  • triciaal
do you have answer choices?
anonymous
  • anonymous
No, just a set number of attempts.
triciaal
  • triciaal
do you have an idea what my sketch is?
anonymous
  • 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
  • anonymous
*got
triciaal
  • triciaal
|dw:1443677500413:dw|
triciaal
  • triciaal
you skipped the leading minus
triciaal
  • triciaal
it said measured clockwise
anonymous
  • anonymous
Shalante said they figured it out. Hold please,
triciaal
  • triciaal
@dan815 what is your take on this?
anonymous
  • 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
  • anonymous
By the way, I am pretty active in the physics section, but not engineering though.
anonymous
  • anonymous
If two objects are traveling in opposite we would add them up. Just like how cars move very fast in opposite direction.
anonymous
  • 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
  • anonymous
Wait, nevermind. I just realized my mistake.
triciaal
  • 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
  • anonymous
I wrote the wrong value for that variable on my physical paper, so no wonder I've been getting this wrong.
anonymous
  • 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
  • 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
  • anonymous
Yea, no problem. Solving another difficult physics question right now.
anonymous
  • anonymous
Thanks for the medals
anonymous
  • anonymous
medal*

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