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Projectile motion finding initial speed of a ball pushed off a table

  1. Sep 28, 2008 #1
    A ball is projected horizontally from the edge
    of a table that is 0.823 m high, and it strikes
    the floor at a point 1.44 m from the base of
    the table.
    The acceleration of gravity is 9.8 m/s2 .
    What is the initial speed of the ball? An-
    swer in units of m/s.


    the table is .823 m high.
    x = xo + vot+ .5at^2
    x = .823 , xo = 0, vot = 0 (vertical speed is 0), a = 9.8 m/s2 and you solve for t
    .823 m = .5(9.8 m/s2)t^2
    t^2 = .1679 s
    t = .41 sec

    to find the intial speed you know the distance and the time
    x = xo +vot+.5at^2
    1.44 m = 0 m + vo(.41 s) + .5(9.8 m/s2)(.41 s)^2
    you get vo = 1.5 m/s

    I tried this but I did something wrong because i enter my hw online and it told me the answer i got was incorrect.
     
    Last edited: Sep 28, 2008
  2. jcsd
  3. Sep 28, 2008 #2

    gabbagabbahey

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    Homework Helper
    Gold Member

    Gravity is directed downward, so why do you have the projectile accelerating horizontally in your calculation for v0?
     
  4. Sep 28, 2008 #3
    Thanks! Your absolutely right that fixed it. I feel kinda dumb now. We just learned the verticle horizontal stuff so i'm still a bit confused and get flipped around. thanks again :)
     
  5. Oct 1, 2008 #4
    use x=vicostheta t
    anser is 3.5
     
  6. Oct 1, 2008 #5
    First solve for time: t=sqrt of 2*height/g

    t=sqrt of 2(.823)/9.8 = .43 secs. This is how long it takes to fall vertically, the horizontal and vertical are tied through time.

    Given that the object travels 1.44m horizontally in .43 sec we can divide 1.44m by .43 secs to get 3.348 m/s or 3.5 m/s.

    The Trig way works too! but if you don't have an angle use this method.
     
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