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Algebra-Quadratic Applications

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A diver on a platform 40 feet in height jumps upward with an initial velocity of 5 feet per sec. His height in h feet after t seconds is given by the function h=-16t^2+5t+50. What is his maximum height? How long will it take him to reach the surface of the water?

  • Algebra-Quadratic Applications - ,

    There is a discrepancy in your problem.
    You state the platform is 40 feet high, yet the equation suggests an initial height of 50 feet.

    You would find the vertex of the downwards parabola.
    An easy way is to find the x of the vertex by -b/2a
    = -5/(-32) = 5/32 seconds
    Now sub that back in to find the height when t = 5/32

    for the last part,
    set -16t^2 + 5t + 50 = 0
    and solve for t using the quadratic formula.

  • Algebra-Quadratic Applications - ,

    I assume you have not covered calculus which would make this easy.
    Without calculus you must complete the square to find the vertex of the parabola.
    h=-16t^2+5t+50
    -h = 16 t^2 - 5t - 50
    50 - h = 16 t^2 - 5 t
    50/16 -h/16 = t^2 -(5/16) t
    50/16 - h/16 + 25/1024 = t^2 -5/16 t + 25/1024

    3225/1024 - h/16 = (t-5/32)^2
    we are at the top when t = 5/32
    and h = 3225/64 = 50.4

    Now what is t when h = 0?
    = -16 t^2 +5 t + 50
    t = [ -5 +/- sqrt (25 + 3200) ]/-32
    t = [-5 + 56.8 ]/-32 = -1.62
    ignore that solution, negative time was before we started
    t = [-5 - 56.8]/-32 = 1.93 seconds

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