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?
A. 6
B. 86
C. 134
D 150
E. 214
Equation, rephrased:
h = 150 - 16(t-3)².
To MAXIMIZE the value of h, we need to MINIMIZE the value SUBTRACTED from 150: 16(t-3)².
Since (t-3)² cannot be negative, the smallest possible value of 16(t-3)² is 0.
16(t-3)² = 0 when t=3.
Thus, the maximum height occurs when t=3.
Two seconds later, t=5.
When t=5, h = 150 - 16(5-3)² = 150-64 = 86.
The correct answer is
B.
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