SOLUTION: A student throws her math instructor over a building. His height above the ground after t seconds is given by the equation: h(t)=-16.1t^2+6t+75 How tall is the building?

Question 618504: A student throws her math instructor over a building. His height above the ground after t seconds is given by the equation:
h(t)=-16.1t^2+6t+75
How tall is the building?
Found 2 solutions by swincher4391, stanbon:
Answer by swincher4391(1107) (Show Source): You can put this solution on YOUR website!
Consider t=0. At this time the math instructor is at the top of the building not yet thrown. So following this logic, h(0) must be the height of the instructor above the ground at time 0. This is the height of the building. It is trivial to see that h(0) = 75.
Thus, the building is 75 (units?).
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
h(t)=-16.1t^2+6t+75
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maximum height occurs when t = -b/(2a) = -6/(2*-16.1) = 0.1863
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The maximum height is h(0.1863) = -16(0.1863)^+6(0.1863)+75 = 75.56 ft.
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Cheers,
Stan H.
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