An object thrown directly upward is at a height of h feet after t seconds, where h = -16 (t - 3)2 +
150. At what height, in feet, is the object 2 seconds after it reaches its maximum height?
Max height: t=3, h=150. You know that anything^2 ((t-3)^2) is positive. A negative number (-16) times a positive number ((t-3)^2) must be negative. So to maximize 'h' you want to minimize '-16(t-3)^2'.
2 seconds after max height: t=5, h=86.
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