Algebra

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If a ball is thrown vertically upward from a height of 56ft. above ground with an initial velocity of 40ft. per second, then the height of the ball above ground t seconds after it is thrown is given by f(t)=-16t^2 + 40t +56. How many seconds will elapse after the ball is thrown before it hits the ground?

I need step by step explaination please. I know the answer is 3.5 seconds.

  • Algebra -

    Well, you have a function. If f(t) = the height, when does f=0?

    In other words, just solve the dang quadratic!

    -16t^2 + 40t +56 = 0
    t = -1 and 3.5

    See whether you can pick the proper one.

  • Algebra -

    okay, so I solved as follows:
    -16t+40t+56=0
    (4t+4)(-4t+14)
    4t+4=0 and -4t+14=0
    4t=4 and -4t=-14
    4t=4/4 and -4t= -14/-4
    = t=1 and t = 7/2

    Now, I am lost at what to do from here.

  • Algebra -

    duh, I got it. you divide 7/2 to equal 3.5

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