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object thrown directly upward

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by maihuna » Sun Sep 13, 2009 6:10 am
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
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by ssmiles08 » Sun Sep 13, 2009 6:19 am
The ball reaches its maximum height at 3 sec. Anything between 0-2 seconds will give you a smaller value because you would be subtracting from 150. The same can be said for a number slightly > 3 seconds.

3 + 2 = 5 sec.

h = -16 (5 - 3)^2 + 150 = -64 + 150

= 86.

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by prindaroy » Sun Sep 13, 2009 3:00 pm
The above poster is wrong;

when t = 1

h = -16(1-3)2 + 150 = 214

when t = 0

h = 96 + 150 = 246

so, the question is a little messed up. Because there is nothing that says it will reach a max. height at 3. It could reach it at 1 second.
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