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megd

  • 3 years ago

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

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  1. TuringTest
    • 3 years ago
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    this problem is word-for word for sure? like "below floor level" and all that weird-sounding stuff?

  2. megd
    • 3 years ago
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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

  3. FreeTrader
    • 3 years ago
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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.

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

  5. FreeTrader
    • 3 years ago
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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.

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

  7. FreeTrader
    • 3 years ago
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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?

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

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

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

  11. TuringTest
    • 3 years ago
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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

  12. TuringTest
    • 3 years ago
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    my calculator has it pointing upwards

  13. megd
    • 3 years ago
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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

  14. megd
    • 3 years ago
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    what did you use as the equations then?

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

  16. TuringTest
    • 3 years ago
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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...

  17. megd
    • 3 years ago
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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..

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

  19. TuringTest
    • 3 years ago
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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

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

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

  22. TuringTest
    • 3 years ago
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    later :)

  23. megd
    • 3 years ago
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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?

  24. TuringTest
    • 3 years ago
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    what table?

  25. megd
    • 3 years ago
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    hit 2nd and graph

  26. megd
    • 3 years ago
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    i would get 1.3 seconds

  27. megd
    • 3 years ago
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    sorry 1.13

  28. TuringTest
    • 3 years ago
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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

  29. megd
    • 3 years ago
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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

  30. TuringTest
    • 3 years ago
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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.

  31. phi
    • 3 years ago
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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

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

  33. phi
    • 3 years ago
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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

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

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

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

  37. TuringTest
    • 3 years ago
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    which part?

  38. megd
    • 3 years ago
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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?

  39. TuringTest
    • 3 years ago
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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?

  40. TuringTest
    • 3 years ago
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    by r I mean the distance travelled

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

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

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

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

  45. TuringTest
    • 3 years ago
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    oh, I forgot conversions...

  46. megd
    • 3 years ago
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    i did that at first too...

  47. phi
    • 3 years ago
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    But it does hit the mark.

  48. megd
    • 3 years ago
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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?

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

  50. megd
    • 3 years ago
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    its okay thanks!

  51. phi
    • 3 years ago
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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

  52. megd
    • 3 years ago
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    and how exactly would you set that up?

  53. phi
    • 3 years ago
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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)

  54. phi
    • 3 years ago
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    plug into the first equation to find v

  55. megd
    • 3 years ago
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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

  56. phi
    • 3 years ago
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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)

  57. megd
    • 3 years ago
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    ok

  58. megd
    • 3 years ago
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    so what would the answer for that be?

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