## 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 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?

1. 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.

2. anonymous

Is anybody there?

3. anonymous

Hello?

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. triciaal

|dw:1443676636690:dw|

8. triciaal

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

9. triciaal

10. anonymous

No, just a set number of attempts.

11. triciaal

do you have an idea what my sketch is?

12. 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.

13. anonymous

*got

14. triciaal

|dw:1443677500413:dw|

15. triciaal

16. triciaal

it said measured clockwise

17. anonymous

Shalante said they figured it out. Hold please,

18. triciaal

@dan815 what is your take on this?

19. 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

20. anonymous

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

21. anonymous

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

22. 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?

23. anonymous

Wait, nevermind. I just realized my mistake.

24. 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

25. anonymous

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

26. 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.

27. 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.

28. anonymous

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

29. anonymous

Thanks for the medals

30. anonymous

medal*