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Non Symetrical Parabola Question

  1. Sep 14, 2008 #1
    1. The problem statement, all variables and given/known data
    In a local bar , a customer slides an empty beer mug down the counter for a refill. The bartender is just deciding to go home and rethink his life. He does not see the mug, which slides off the counter and strikes the floor 1.40m from the base of the counter. If the height of the counter is .860m. (a) with what velocity did the mug leave the counter and (b) what was the direction of the mug's velocity just before it hit the floor?


    2. Relevant equations
    (a) I got the correct answer for this by using the equation
    yf= yi + vyit-1/2gt2
    I got 3.3 m/s
    (b) I would assume that since I know the height of the bar at .860m and the distance
    traveled of 1.4m


    3. The attempt at a solution
    (b) The angle would make sense to me to be the tangent of opposite over adjcent but
    I get 31.6 degrees, but from the book the answer is -50.9
    Not sure what I'm doing wrong.

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



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 14, 2008 #2

    LowlyPion

    User Avatar
    Homework Helper

    Let's see what is the title of this problem?
    Non Symetrical Parabola Question

    Does that suggest anything about what path the mug takes from bar top to floor? A triangle? Think more expansively.

    What vertical velocity does it achieve? And how long to get there after leaving the bar top?

    vy2/(2*a) = Height
    Height = 1/at2

    Now you have the vertical velocity and you can figure the horizontal velocity from Vx = x/t those are the two components when it hits. (Of course Vx is the speed it leaves the counter.)

    The Root Sum of the Squares then of the two components is the total magnitude and the two components determine direction (angle θ).
     
  4. Sep 14, 2008 #3
    That's begining to make more sense.
    I used VyF=Vyi-gt
    So I get Vyf = 3.3m/s-(9.8m/s2).42s
    Vyf = -.82m/s

    Time I got by using Yf=Yi+Vyi(t)-1/2gt2
    I then solved for t and get .42s

    Not sure where to go from here.
    Kevin
     
  5. Sep 14, 2008 #4
    Scratch the last. I didn't do that right
     
  6. Sep 14, 2008 #5
    I get Vyf to be -7.42 m/s
    and Vyi to be 3.3 m/s
    Are these right?

    Kevin
     
  7. Sep 14, 2008 #6

    LowlyPion

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    Homework Helper

    Answering these questions solves your problem.

    How much time for the mug to drop to the floor?
    Height = 1/2at2
    .86 = 1/2 (9.8) t2
    t = .419 sec

    X Distance= 1.4 m
    Horizontal velocity = 1.4/.419 = 3.34 m/s

    Vertical velocity:
    V2 = 2*a*x
    V2 = 2*9.8*.86 = 4.11 m/s

    Angle θ with the floor:

    tan-1(4.11/3.34) = Angle θ
     
  8. Sep 14, 2008 #7
    Thank you for the help.
    I used the equation Vyf = Vyi -gt and got -7.42 m/s
    Your equation makes alot of sense but I couldn't find that one.
    I'm not sure how to tell when to use which one. Is mine for a symetrical
    parabola.
    Is there an easy way to find the y value on a symetrical parabola. I have a total
    diatance, the disatance at which I need the height, the initial velocity, and the initial
    angle?
    Thanks again,
    Kevin
     
  9. Sep 14, 2008 #8

    LowlyPion

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    Homework Helper

    Basically there are a couple of useful relationships, and it is likely more useful to understand these, because they allow you to partition a problem into pieces and combine what you need to solve for.

    V = a* t
    Acceleration times time is velocity.

    V = x/t
    Definition of velocity

    x= 1/2 at2
    the distance a body at rest falls related to time

    V2 = 2ax
    relates the max velocity from rest under constant acceleration.

    Now using those you can add initial conditions and develop more complex models based on what the question asks. But those are always a good place to start expanding the things you know about a problem as you move toward a solution.

    Taking the more general form of these equations requires you to be careful about which time and which velocity is initial and final and can sometimes create more confusion than using the simpler forms and adding the terms that are relevant.
     
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