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# An object thrown directly upward is at a height of h feet

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Director
Joined: 06 Jan 2008
Posts: 546
An object thrown directly upward is at a height of h feet [#permalink]

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28 Apr 2008, 13:27
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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
Senior Manager
Joined: 19 Apr 2008
Posts: 317

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28 Apr 2008, 14:16
saravalli wrote:
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

can you please put the proper sign * , ^ in the equation of h given ? I can't interprete it correctly.
Director
Joined: 06 Jan 2008
Posts: 546

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28 Apr 2008, 17:16
h = -16 (t - 3)^2 + 150
sorry
Current Student
Joined: 17 Jan 2008
Posts: 585
Location: Ann Arbor, MI
Schools: Ross '12 (MBA/MS)

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28 Apr 2008, 18:11
saravalli wrote:
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

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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Senior Manager
Joined: 06 Aug 2007
Posts: 361

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29 Apr 2008, 17:59
saravalli wrote:
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

Max height is when t = 3 so the first part = 0 and the Max height = 150

After 2 sec after max height is when t = 5
substitute and you will get B!!!
Re: PS - object   [#permalink] 29 Apr 2008, 17:59
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