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Quadratic applications question

  1. Apr 29, 2007 #1
    1. The problem statement, all variables and given/known data
    A projectile is launched upward. Its height above the ground after t seconds is -16t^2 + 320t. After how many seconds in the air will it return to the ground? Solve algebraically.

    2. Relevant equations
    ax^2+bx+c=0 ?
    -b (+/-) (Square root of b^2 -4ac)/2a ?

    3. The attempt at a solution

    t= -320 (+/-) (Square root of 320^2 -4(-16)(0)

    t= -320 (+/-) 320/ -16
    t= {0,20} ?

    If i check it, it doesnt fit.
  2. jcsd
  3. Apr 29, 2007 #2

    Gib Z

    User Avatar
    Homework Helper

    Its simpler this way.


    Divide both sides by -16




    We get the same solutions, check it the long way by paper, maybe your calculators got errors. My solutions are verified by my calculator..
  4. Apr 29, 2007 #3
    20=-16(20^2) + 320(20) ?
    20= -6400+6400?

    20 =/= 0

    0= -16(0^2) + 320(0) ?

    Now, since this is geometry, t(seconds) cannot equal 0 or else the time and distance would be also 0

    If i plug in 20 as t, it ends up as 20=0, which is not true.

    I don't understand how I am right
  5. Apr 29, 2007 #4
    Why not? t=0 is just the point of firing - one of two points described by the parabola where the height is zero. What other value of t allows the equation:

    -16t^2 + 320t = 0

    to equal zero? This:

    Is wrong. Why have you placed 20 on the left hand side?
  6. Apr 29, 2007 #5
    Right.... lol Thanks!!~!~
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