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Problems with Velocity and Acceleration

  1. Feb 22, 2006 #1
    A ball thrown straigh upward from the ground reaches a maximum height of 400ft. What was its initial velocity? Using the formulas:
    the distance travelled is 400ft, and the initial distance was 0. The force of gravity is -32ft/s/s. Also, the velocity of the ball at the top of the height is 0.

    in the equation V=at+Vo, I don't know how to solve it with 2 variables (t and Vo)
    Last edited: Feb 22, 2006
  2. jcsd
  3. Feb 22, 2006 #2
    V=at+Vo ---------------(A)
    X=1/2at^2+Vot+Xo -----------(B)

    ok. couple of things to note

    assume we are measuring height from ground. (i.e. X0 = 0)
    at maximum height, the velocity of the ball becomes zero for a moment just before it turns around. so at the point V = 0

    so equation (A) becomes
    0 = at + V0 => t = - V0/a

    here 'a' is gravitational acceleration. u need to put correct sign as well.

    i.e. a = -(g). that way your time is positve. (time cannot be negative)

    now substitute this in to (B) and solve for V0.
  4. Feb 22, 2006 #3
    thanks but I'm still stucki got Vo^2=12800/-15 and you can't do a square root of a negative :S
  5. Feb 22, 2006 #4
    thank you so much, i got it. except i made Vo=32t instead. :D
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