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Math Help - what is the maximum height...

  1. #1
    Senior Member
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    what is the maximum height...

    ok so this problem seems pretty straight forward.

    a ball is thrown straight up from the ground with an initial velocity of 256 ft/s. the equation h = 256t -16t^2 describes the height the ball can reach in t seconds.

    ok so i take the first derivative and find that t = 8 seconds or in 8 seconds the ball reaches the max height.

    when i plugged in the 8 seconds i got 1024 as the max height in feet, my book says the answer is wrong? ok so where did i mess up?
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  2. #2
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    Re: what is the maximum height...

    Seems correct to me. Even starting with the acceleration function and integrating to velocity you obtain the same answer.
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  3. #3
    MHF Contributor

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    Re: what is the maximum height...

    Because that is a simple quadratic you can also find the maximum height by completing the square:
    h= 256t- 16t^2= -16(t^2- 16t+ 64- 64)= -16(t- 8)^2+ 1024
    Since a square is never negative, -16(t- 8)^2 is never positive and h is never larger than 1024 which is its value when t= 8.
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