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megd Group Title

suppose a molecule of water is propelled from a point 6 inches below floor level at an angle of 60 with an initial velocity of 21.3 feet per second. write modeling equations that describe the position of the molecule in the water stream for any time t

  • 2 years ago
  • 2 years ago

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  1. TuringTest Group Title
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    this problem is word-for word for sure? like "below floor level" and all that weird-sounding stuff?

    • 2 years ago
  2. megd Group Title
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    yes that's what it said word for word i tried plugging this into my calculator: x = 21.3T cos (60) and y = 21.3T sin (60)-16t^2 - 0.5 and the graph goes downward which isn't right

    • 2 years ago
  3. FreeTrader Group Title
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    1) Convert units to be the same. 2) Draw a diagram (on paper) of what's going on. 3) Use trig relationships to figure out missing parts of your diagram. 4) Integrate the velocity formula to find the position function. That should get you pointed in the right direction.

    • 2 years ago
  4. megd Group Title
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    tried graphing it and it points downward which isn't right..

    • 2 years ago
  5. FreeTrader Group Title
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    You stated you tried graphing it with a calculator graphing utility. I suggest you draw it out on paper and begin thinking about it from there.

    • 2 years ago
  6. megd Group Title
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    tried that too and it seems like it should point upward

    • 2 years ago
  7. FreeTrader Group Title
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    Okay, label all the information you have on your diagram. Can you use Trig to find the missing information you need to solve for the position function?

    • 2 years ago
  8. TuringTest Group Title
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    There is no way it should be pointing upward with the angle at 60 deg

    • 2 years ago
  9. megd Group Title
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    well this if for parametric equations so yes you can

    • 2 years ago
  10. megd Group Title
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    not straight upward but to some degree yes it should be

    • 2 years ago
  11. TuringTest Group Title
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    your calculator isn't in radians or anything is it? and what do you mean above? how can the fact that is is parametric change the fact that it angles upwards if initiated at 60 deg

    • 2 years ago
  12. TuringTest Group Title
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    my calculator has it pointing upwards

    • 2 years ago
  13. megd Group Title
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    no its in degree mode. it should be in quadrant 1 and slanted at a 60 degree angle, i mean above the x axis basically, when i graphed it, it appeared below the x axis which cant be right

    • 2 years ago
  14. megd Group Title
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    what did you use as the equations then?

    • 2 years ago
  15. FreeTrader Group Title
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    Are you translating by 6 inches? That would be below the x axis pointed up.

    • 2 years ago
  16. TuringTest Group Title
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    same as you x=21.3Tcos60 y=21.3sin60-16t^2-.5 wait, I did something when I tried to zoom and it changed downward, weird...

    • 2 years ago
  17. megd Group Title
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    i put 0.5 so thats like 6 inches. yeah this doesnt make much sense....the next question is how long is the water molecule in the air before coming down to floor level..

    • 2 years ago
  18. megd Group Title
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    but if you look at the table it appears to be positive for just a little bit

    • 2 years ago
  19. TuringTest Group Title
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    for the next part we should be solving\[h(t)=21.3t \sin(60)-16t^2-0.5=0\]to find the time in the air

    • 2 years ago
  20. TuringTest Group Title
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    of course we will get two answers and we want the larger one

    • 2 years ago
  21. FreeTrader Group Title
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    Best wishes with this. My time's up for today. Bye guys.

    • 2 years ago
  22. TuringTest Group Title
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    later :)

    • 2 years ago
  23. megd Group Title
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    but couldnt you just look at the table? i would get between 1.1 and 1.15 if i do it that way?

    • 2 years ago
  24. TuringTest Group Title
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    what table?

    • 2 years ago
  25. megd Group Title
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    hit 2nd and graph

    • 2 years ago
  26. megd Group Title
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    i would get 1.3 seconds

    • 2 years ago
  27. megd Group Title
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    sorry 1.13

    • 2 years ago
  28. TuringTest Group Title
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    I keep getting different answers... I must be tired. did I put this into wolfram right? http://www.wolframalpha.com/input/?i=t%3D%28-21.3sin60%2Bsqrt%28%2821.3sin60%29%5E2-4*-16*-.5%29%29%29%2F%282*-16%29

    • 2 years ago
  29. megd Group Title
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    im not really familiar with that site....but the next part was to figure out the maximum height and i got 4.816 feet from the table which doesnt make sense either

    • 2 years ago
  30. TuringTest Group Title
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    look at the link you should get to learn the site, it is very useful for this type of thing you can see if the formula entered is the same as yours I have to go, sorry. Good luck though.

    • 2 years ago
  31. phi Group Title
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    Here's wolfram's plot http://www.wolframalpha.com/input/?i=plot+x+%3D+21.3*t*cos+%2860%29+%2C+y%3D+21.3*t*sin%2860%29+-+16t%5E2+-0.5

    • 2 years ago
  32. TuringTest Group Title
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    thanks phi, I couldn't figure how to put parametric equations in wolfram. "plot", huh?

    • 2 years ago
  33. phi Group Title
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    and here's t for y=0 http://www.wolframalpha.com/input/?i=solve++0%3D+21.3*t*sin%2860%29+-+16t%5E2+-0.5

    • 2 years ago
  34. megd Group Title
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    oh wow yeah my graph looks nothing like that..dont know why

    • 2 years ago
  35. TuringTest Group Title
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    oh, I forgot to use the negative root to make for the larger time, because the denominator is negative...

    • 2 years ago
  36. megd Group Title
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    well i guess it sort of does match my answers, confused about the next part though

    • 2 years ago
  37. TuringTest Group Title
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    which part?

    • 2 years ago
  38. megd Group Title
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    d.) it is 12 feet from the point at which the water is propelled to the center of the receptacle in the floor where it needs to land. if the receptacle is circular with a diameter of 3 inches, will the water molecule land in the receptacle?

    • 2 years ago
  39. TuringTest Group Title
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    well you need to find r you have the time t of travel in the air, so the distance it travels is v_x=21.3cos(60)t=21.3cos(60)(1.125) now is this x within \[12\pm1.5\]where 1.5 is the radius of the receptacle?

    • 2 years ago
  40. TuringTest Group Title
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    by r I mean the distance travelled

    • 2 years ago
  41. TuringTest Group Title
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    *should be r=(v_x)t=21.3cos(60)t=21.3cos(60)(1.125)

    • 2 years ago
  42. megd Group Title
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    okay based on that i think it will make it to the receptacle...

    • 2 years ago
  43. phi Group Title
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    double check it's 1.5 inches so 1.5/12 feet

    • 2 years ago
  44. TuringTest Group Title
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    right, because \[21.3\cos(60)(1.125)=11.98\] which is in \[12\pm1.5\]

    • 2 years ago
  45. TuringTest Group Title
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    oh, I forgot conversions...

    • 2 years ago
  46. megd Group Title
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    i did that at first too...

    • 2 years ago
  47. phi Group Title
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    But it does hit the mark.

    • 2 years ago
  48. megd Group Title
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    okay good, so last thing is e.) an arc of water needs to land in a 3-inch circular receptacle in the floor whose center is 15 feet from the point at which the water is propelled. suppose the water is propelled from a point 6 inches below floor level at an angle of 55. at approximately what intitial velocity must the water be propelled in order to land in the receptacle?

    • 2 years ago
  49. TuringTest Group Title
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    sorry, I'm going to bed, hopefully phi can help you. g'night!

    • 2 years ago
  50. megd Group Title
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    its okay thanks!

    • 2 years ago
  51. phi Group Title
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    2 equations and 2 unknowns. I'd use the x one to find v in terms of t. sub into the y equation, and solve for t when y=0

    • 2 years ago
  52. megd Group Title
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    and how exactly would you set that up?

    • 2 years ago
  53. phi Group Title
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    x= v t cos(55), with x= 15 so 15= v t cos(55) v= 15/(t cos(55) y= v t sin(55) - 16 t^2 -0.5, with v as above. solve for t when y= 0 ( the water hits the floor)

    • 2 years ago
  54. phi Group Title
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    plug into the first equation to find v

    • 2 years ago
  55. megd Group Title
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    sorry but how do you solve for t when y = 0...im at 0 = (15/t cos (55)) t sin (55) - 16t^2 - 0.5

    • 2 years ago
  56. phi Group Title
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    For starter's, the t cancels in the first term. The sin, cos are constants so it's essentially a t^2 = b for t= sqrt(b/a)

    • 2 years ago
  57. megd Group Title
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    ok

    • 2 years ago
  58. megd Group Title
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    so what would the answer for that be?

    • 2 years ago
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