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Homework Help: Equation for initial velocity

  1. Sep 8, 2006 #1
    Problem:
    A ball is thrown vertically upward, which is the positive direction. A little while later it returns to its point of release. The ball is in the air for a total time of 10 s. Note: Near the earth's surface, g is approximately 9.80 m /s2.

    (a1) What is the algebraic expression for the initial velocity v0 of the ball? Express your answer in terms of the ball's displacement y, its acceleration a in the vertical direction, and the elapsed time t.

    ---------
    is this correct?:
    Equation:
    v0=(y-.5at^2)/t

    v0=(0-490)/10
    v0=(-490)/10
    v0=-49
     
  2. jcsd
  3. Sep 8, 2006 #2

    tony873004

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    "A ball is thrown vertically uwpard which is the positive direction"

    And you get -49, which since its negative, is the down direction. Will a ball thrown towards the ground last 10 seconds in the air? Only if you throw it from the roof of a tall building.
     
  4. Sep 8, 2006 #3
    You seem to have used the formula y=ut+0.5a(t^2), transposed it and obtained u=(y/t)-0.5at, so as long as you have your positive and negative directions right you've done it right, i think.
     
  5. Sep 8, 2006 #4
    one thing to think about, if the positive direction is upwards, how could the ball travel up when it has a negative initial velocity?
     
  6. Sep 8, 2006 #5

    radou

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    The vertical 'throw' equation for the y-direction is [tex]y(t)\vec{j}=v_{0}t\vec{j}-0.5gt^2\vec{j}[/tex], which is equal to y(t) = v0*t - 0.5*g*t^2. Actually, I don't see any problems with the directions; the initial velocity v0 is 'positive', and gravitiy is acting in the 'negative' direction all the time. I don't see how you got a negative velocity from your equation. v0 equals 49 [m/s].
     
  7. Sep 8, 2006 #6
    So my answer is correct? just instead of negative its positive?
    i got negative because, the displacement is 0,
    v0=(y-.5at^2)/t
    v0=(0-.5(9.8)(10^2))/10
    v0=(0-490)/10
    v0=-490/10
    v0=-49

    However when i use the equation radou gives, i get the positive 49... so i think its just the eqution i used..... Thanks guys!
     
  8. Sep 8, 2006 #7

    nrqed

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    Your equation is correct, it's just that the acceleration due to gravity is [itex] a_y = -g = -9.80 m/s^2 [/itex]. (assuming the positive y axis pointing upward). You used a=+g which is the problem.

    Hope this helps
     
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