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2 equations 2 unknowns

  1. Dec 6, 2009 #1
    1. The problem statement, all variables and given/known data
    Ok forgive me as an engineering student but this problem should be easier than it seems. The Problem: It is observed that the skier leaves the ramp A at an angle (Theata=25 degrees) with the horizontal. If he strikes the ground at point B, determine his initial speed,V, and the time of flight,t.


    2. Relevant equations

    I have used:

    s=vt
    s=s+vt+1/2at^2

    The dimensions needed are correct and are 100m down the slope alligned, the ramp he leaves from is 4m high and the ramp is angled at a 3,4,5 triangle.

    This problem is 12-110 from the Hibbeler Dynamics 12th edition
    3. The attempt at a solution

    I have setup the equations as

    (1) 100(4/5)=Vcos(25)t
    (2) -4-100(3/5)=0+Vsin(25)t+(1/2)(-9.81)t^2

    I have tried solving (1) for t and plugging in into (2) but come out with a strange decimal and also solving (1) for V and plugging into (2) I can't seem to get it.

    I know the answers are supposed to be: V=19.4m/s t=4.54s

    Please help!!!
     
  2. jcsd
  3. Dec 8, 2009 #2

    berkeman

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    Staff: Mentor

    A 3-4-5 right triangle does not give you a takeoff angle of 25 degrees.
     
  4. Dec 8, 2009 #3
    Sorry about the confusion and I am aware that a 3-4-5 triangle does not make a 25 degree takeoff. Here is a free body diagram to help describe. The equations are correct btw, I just can't solve the system. Thanks for the help.
     

    Attached Files:

  5. Dec 8, 2009 #4

    berkeman

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    Ah, that helps. Where did the "-4" come from?


    EDIT -- Oh wait, I see the ramp 4m offset in the figure now.
     
    Last edited: Dec 8, 2009
  6. Dec 8, 2009 #5

    berkeman

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    So as you said, you have two equations and two unknowns. How can you go about solving for V and t?

    [tex]100(\frac{4}{5})=Vcos(25)t [/tex]
    [tex]-4-100(\frac{3}{5})=0+Vsin(25)t+(1/2)(-9.81)t^2 [/tex]
     
  7. Dec 8, 2009 #6
    Yeah I need to solve for V and t.
     
  8. Dec 8, 2009 #7

    berkeman

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    So have at it! What would be a good way to start?
     
  9. Dec 8, 2009 #8
    !?!?!?! I just ran through it again and it worked out. Unbelievable. I guess what thy say about walking away from the problem and coming back to it later really works. Im gonna post my work. Thanks for your help.
     
  10. Dec 8, 2009 #9

    berkeman

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    Staff: Mentor

    Great! Good job.
     
  11. Dec 8, 2009 #10
    Here is the work. It's a bit sloppy because I ran through it. Please excuse the mess.
     

    Attached Files:

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