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Ball projection problem help

  1. Sep 13, 2009 #1
    1. The problem statement, all variables and given/known data

    A ball is projected horizontally from the edge of a table that is 0.443 m high, and it strikes the floor at a point 1.84 m from the base of the table.

    The acceleration of gravity is 9.8 m/s^2


    2. Relevant equations

    a) What is the initial speed of the ball? Answer in units of m/s

    b) How high is the ball above the floor when its velocity vector makes a -18.9486 degree angle with the horizontal? Answer in units of m.


    3. The attempt at a solution

    a) i got it right, and the answer is 6.13

    b) this is what i have trouble on. I started by finding Vy using (6.13)(tan -18.9486)=Vy and i got: Vy=-2.10 I then used the formula Vf = Vo + a*t to get time. I got t=-.21 s
    Next i used d = Vot + 0.5a*t^2 to solve for distance and i got d= .22

    Finally i did 0.443-0.22 and i got 0.22. The answer is not .22.

    I need help please. What did i do wrong?
     
  2. jcsd
  3. Sep 14, 2009 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi Sealy! Welcome to PF! :smile:
    (probably a rounding error, but anyway …)

    there are three standard https://www.physicsforums.com/library.php?do=view_item&itemid=204" equations …

    why not use v2 = u2 + 2as ? :wink:
     
    Last edited by a moderator: Apr 24, 2017
  4. Sep 14, 2009 #3
    I tried to redo the problem without rounding until the end and i still got 0.22. I even tried 0.21 and 0.23. I sent a copy of my work and my solution to my teacher today so i'm waiting for her response.

    I really think i used the right method.
     
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