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Motion in two or three dimensions

  1. Sep 1, 2011 #1
    Hi. I am lost in this problem. Could someone tell me what to do. How to start with the problem?
     

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  2. jcsd
  3. Sep 1, 2011 #2

    PeterO

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    Start by considering the vertical and horizontal components of the initial speed.
     
  4. Sep 1, 2011 #3
    Which equation do I need to use?
     
  5. Sep 1, 2011 #4

    PeterO

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    One of the components is Vsin(53) the other is Vcos(53)
     
  6. Sep 1, 2011 #5
    Vx = V0*cos(53,0)

    Vy = V0*sin(53,0)
     
  7. Sep 1, 2011 #6

    PeterO

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    Now you have to use them.

    One will tell you how long it takes to get across the river, the other will let you calculate how long it takes the car to go up in the iar, then come back down again to the height of the far bank, and those two times have to be the same.
     
  8. Sep 1, 2011 #7
    I have just seen in my book, that the last equations needs to be:

    Vy = V0*sin(53,0) - g*t

    Is that true?

    So I get these two equation, that I need to use as you said:

    Vx = V0*cos(53,0)

    Vy = V0*sin(53,0) - g*t
     
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