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Motion in 2D Help

  1. Jun 30, 2007 #1
    Motion in 2D!!! Help

    1. The problem statement, all variables and given/known data

    A stone is thrown upwards from the top of a building at an angle of 30 degree to the horizontal and with an initial speed of 20 m/s. The point of release is 45 m above the ground. (a) How long does it take the stone to hit the ground?

    2. Relevant equations Solved

    Vo = 20 m/s
    Voy = Vo*Sin[tex]\theta[/tex] => 10.0 m/s
    A = -g => -9.8 m/s^2

    [tex]\Delta[/tex]y = Voy*t - 1/2(g)(t)^2

    3. The attempt at a solution

    I solved all the required equations as seen above to solve the equation but i need to find time in seconds and i dont know how ???? HELPPPPP

    [tex]\Delta[/tex]y = Voy*t - 1/2(g)(t)^2
    -45m = (10m/s)t - (0.5)(9.8m/s^2)(t^2)
  2. jcsd
  3. Jun 30, 2007 #2
    You have to use the quadratic forumla.

    t= [-b +/- (b^2 - 4ac)^(1/2)] / (2a)

    This corresponds to an equation in the form of 0 = at^2 + bt + (some constant, in your case 45 since you would add it from the left to the right side of the equation.)

    When you solve for t, you'll get two answers since you must do both plus and minus. If one of the two is negative, that is not the answer. Use the positive time.
    Last edited: Jun 30, 2007
  4. Jun 30, 2007 #3
    Thanks got the answer !!!
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