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Homework Help: Final velocity need help

  1. Feb 14, 2010 #1
    final velocity...need help!!

    1. The problem statement, all variables and given/known data

    An object is to be thrown through a window 3.25 m high a distance 8 m away. It is launched at an angle 75 degree above the horizontal.

    The initial launch speed at the specified angle which will allow the projectile to pass through the window is 13.3 m/s.

    What is the final speed of the object (that is, the speed at which it flies through the window)?
    Y= height
    Yo= initial height
    Voy= initial speed
    Vf= final velocity
    g= gravity
    t= time

    2. Relevant equations
    1)Y= Yo+Voy*t-(1/2)gt^2
    2)Vf= Vo+at


    3. The attempt at a solution

    By using the y= yo+Voy*t-(1/2)gt^2 formula,
    = 3.25+ (13.3sin75)t-(1/2)(-9.81)t^2
    I found the time t=2.85 s

    But then the answer i got by using the formula Vf= Vo+at
    = 13.3+ (-9.81)(2.85)
    = -14.7 m/s
    IS wrong!!What is my mistake??
     
  2. jcsd
  3. Feb 14, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi theoronhalde ! Welcome to PF! :smile:

    (try using the X2 and X2 tags just above the Reply box :wink:)
    Vfy = Voy + at :wink:
     
  4. Feb 14, 2010 #3
    Re: final velocity...need help!!

    hi!! thx a lot!!

    Not sure i understand your answer? Do i have to use the formula you wrote instead of the one i was using to fnd the final velocty? I'm a bit confuse...
     
    Last edited: Feb 14, 2010
  5. Feb 14, 2010 #4

    tiny-tim

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    Yes … you have to deal with the y-components separately. :smile:

    (the x-component of velocity, of course is constant :wink:)
     
  6. Feb 14, 2010 #5
    Re: final velocity...need help!!

    ok s it's going to be:

    Vfy= Voy+at
    = 13.3sin75+ (-9.81)(2.85)
    = -15.1 m/s

    But how come the horizontal velocity be constant? And am I suppose to use
    V=Sqrt((Vox^2)+(Vfy^2)) to find that final speed?
     
  7. Feb 14, 2010 #6

    tiny-tim

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    The horizontal acceleration is zero, so the horizontal velocity must be constant.

    And yes, use Pythagoras. :smile:
     
  8. Feb 14, 2010 #7
    Re: final velocity...need help!!

    oh ok....i undestand what you are saying but i still end up with a wrong answer...here is what i did:

    Vfx= Vox + at (a=0)
    =13.3cos75

    Vfy= Voy+at
    = 13.3sin75 + (9.81)(2.85)
    = -15.1


    Vf= SQRT( Vfy^2 + Vfx^2)
    = 15.5 m/s

    The applet says that the answer is wrong :S ??
     
  9. Feb 14, 2010 #8

    tiny-tim

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    I haven't checked how you got t using Vy, but when I do it using Vx, I don't get 2.85.
     
  10. Feb 14, 2010 #9
    Re: final velocity...need help!!

    oh...i'm so lost once again. I tought we should get the time By using:
    y= yo+Voy*t-(1/2)gt^2 formula,
    = 3.25+ (13.3sin75)t-(1/2)(-9.81)t^2
    If not can you explain me how to find the time and how to get the components Vfy and Vfx.
     
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