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2d kinematics question

  1. Aug 30, 2015 #1
    1. The problem statement, all variables and given/known data
    A 30 degree inline sits on a 1.1 meter high table. A ball rolls of the incline with a velocity of 2m/s. How far does the ball travel acoss the room before reaching the floor?


    2. Relevant equations
    deltax= vot + 1/2at^2


    3. The attempt at a solution

    http:///Users/abhi18/Downloads/IMG_8305.JPG [Broken]http://postimg.org/image/albcq37in/
    The attempt is here

    http://postimg(DOT)org/image/albcq37in/ [Broken]
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Aug 30, 2015 #2

    BvU

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    Hello Scholar, welcome to PF :smile: !

    I can't make your image appear, there's something wrong with the link. Never mind, PF prefers with great emphasis that you post your working by typing it in (there's a zillion of pictures in existing threads that can't be seen any more).

    You mention one releveant equation for motion in the x-direction with an acceleration. What acceleration is there in the x-direction after the ball has left the incline ?
    And what about the y-direction ?
     
  4. Aug 30, 2015 #3
    Hi! The link will appear if you change the (DOT) to .

    Please see the picture because it would be very hard for me to type all the calculations.

    I would greatly appreciate it!
     
  5. Aug 30, 2015 #4

    haruspex

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    I see you obtained a quadratic for t but I do not see a solution for it.
    Neither do I see an equation for ##\Delta x##.
     
  6. Aug 30, 2015 #5

    BvU

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    Done that. What's the argument for replacing the dot by "(DOT)" ? I mean, the "(AT)" instead of @ suppresses some of the junk mail, but what havoc does the (DOT) prevent ?

    Anyway, I also disagree with
    The PF tools are just fine for these few equations.

    You have the answer on the same page as the question ! Comfortable ! So why bring in ##\Delta x= v_0 t + {1\over 2} at^2 ## ?

    Please render the quadratic expression they mention; we'll take it from there.
     
  7. Aug 30, 2015 #6
    (-4.9t^2)+t-1.1

    Using quadratic formula I obtained -.383s and 0.586s. I was confused by the result so to avoid quadratic I tried to find Vfy to avoid the quadratic. So I used Vfy^2=Vo^2+2adeltaY.
    Vf^2=1+(2*-9.8*-1.1)=22.56
    Vf^2= -4.750m/s I used negative because the object is falling.
    Vf=Vo+at
    (-4.750-1)/(-9.8)=t
    t=0.586s
    The book says t=0.383s
    Either the work book is wrong or I am wrong.
     
  8. Aug 30, 2015 #7

    BvU

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    Ok, you're making an effort. Note the nice buttons in the green bar for squares and subscripts ... :rolleyes: and funny faces :smile:

    Words fail me ! You too apparently, because (-4.9t^2)+t-1.1 is very very terse. Does it mean you calculate ##
    \Delta y= v_0 t + {1\over 2} at^2##, in other words you want this to end up as zero ? My guess is the initial velocity is negative and ##\Delta y## is negative too; how does that show up in your expression ?
     
  9. Aug 30, 2015 #8
    I solved it. Thank you. Here is the solution. IMG_8307.jpg
     
  10. Aug 30, 2015 #9

    BvU

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    Sigh ...
     
  11. Aug 30, 2015 #10
    Just because I didn't use the tools you suggested I am incompetent?
     
  12. Aug 30, 2015 #11

    BvU

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    By no means you are incompetent ! You solved it, in a way with which you are comfortable. The sigh was only because for others your ways are hard to follow. Not to mention the pain in the neck from leaning over 90 degrees in order to read it ... :smile:

    (but you rotated since I posted, thanks ! :wink: )
     
    Last edited: Aug 31, 2015
  13. Aug 30, 2015 #12
    I fixed the picture! :wink:.

    Anyways thanks something you said made me realize my error. Please understand this is my first post ever on this site so maybe I wasn't very clear.
     
  14. Aug 30, 2015 #13

    BvU

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    You'll be fine. They tried to teach me to work neatly and failed. But I have to admit it works better in the long run.
     
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