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(h)t=-4.9t^2+98t+8 Find the max height?

  • math -

    are you taking Calculus ?

    h'(t) = -9.8t + 98 = 0
    t = 10
    h(10) = -4.9(100) + 98(10) + 8 = 498


    complete the square

    h(t) = -4.9(t^2 - 20t + 100 - 100) + 8
    = -4.9( (t-10)^2 - 100) + 8
    = -4.9(T-10)^2 + 490+8
    = -4.9(t-10)^2 +498

    vertex is (10, 498) , so max height is 498

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