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anonymous
 one year ago
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 20km/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
 one year ago
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 20km/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?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And 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
 one year ago
Best ResponseYou've already chosen the best response.0@Shalante , I notice you're reading my question. Could you perchance be able to help me out?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Nevermind, just saw your message. Will wait for further instruction.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Again, not looking for answers, here. Just want an idea of how to approach this.

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1dw:1443676636690:dw

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1vertical angles are equal  tan theta =  tan ^1 (6.6/20)

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1do you have answer choices?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, just a set number of attempts.

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1do you have an idea what my sketch is?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0From 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.

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1dw:1443677500413:dw

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1you skipped the leading minus

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1it said measured clockwise

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Shalante said they figured it out. Hold please,

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1@dan815 what is your take on this?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If 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
 one year ago
Best ResponseYou've already chosen the best response.0By the way, I am pretty active in the physics section, but not engineering though.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If two objects are traveling in opposite we would add them up. Just like how cars move very fast in opposite direction.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Where'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
 one year ago
Best ResponseYou've already chosen the best response.0Wait, nevermind. I just realized my mistake.

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1@Shalante thanks I forgot to adjust for the man's velocity @The_Silent_One the raindrop (vertical) was 6.6 the wind was 20

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I wrote the wrong value for that variable on my physical paper, so no wonder I've been getting this wrong.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No 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
 one year ago
Best ResponseYou've already chosen the best response.0Okay, 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
 one year ago
Best ResponseYou've already chosen the best response.0Yea, no problem. Solving another difficult physics question right now.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for the medals
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