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Velocity HELP

  1. Feb 15, 2007 #1
    velocity !!! HELP

    1:you have a lope with an angle of 15 degrees from the horizontal plane. You throw an object with an angle of 60 degrees from the horizontal plane up the slope and the object hits the ground 25m from the starting point. . What is start velocity?

    This is what I’ve done:

    Y = 25m *sin(15)=6,47m
    X= 25m*cos(15)=24,15m
    V0y=sqrt(2*g*y)=11,26m/s
    .t= voy/g=1,15
    V0x=x/t=21m/s
    This gives me this result
    V0= sqrt(v0x^2+voy^2)=23,8 m/s

    But when I check the answere I get the wrong angle?

    Tan(ß)=v0y/v0x -> ß=28,28

    What am I doing wrong ????:cry:
     
  2. jcsd
  3. Feb 15, 2007 #2

    radou

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    The distance in the x direction equals 25 m, perhaps?
     
  4. Feb 15, 2007 #3
    its a slope , why 25 m?
     
  5. Feb 15, 2007 #4

    JK423

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    Gold Member

    "t= voy/g=1,15" ?

    What is that?

    *Wrong comment
     
    Last edited: Feb 15, 2007
  6. Feb 15, 2007 #5
    from the equation vy= v0y-gt, I find the time it takes : 1,15 s
    vy= 0 when it hits the grownd
     
  7. Feb 16, 2007 #6
  8. Feb 16, 2007 #7

    hage567

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    "vy= 0 when it hits the grownd"

    No, it doesn't.
     
  9. Feb 16, 2007 #8

    radou

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    So, it's 25 m along the slope?
     
  10. Feb 16, 2007 #9
    ive solved it, by using
    v0x= v0 *cos(60)
    v0y=v0 *sin(60)
    x= cos(15) *25
    y= sin(15)*25
    and used :
    y=v0y*t-1/2(gt^2)
     
  11. Feb 16, 2007 #10
    ive solved it, by using
    v0x= v0 *cos(60)
    v0y=v0 *sin(60)
    x= cos(15) *25
    y= sin(15)*25
    and used :
    y=v0y*t-1/2(gt^2)

    ;)
     
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