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

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

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30 Sep 2007, 00:15
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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

VP
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Re: PS - SET 26 Q21 [#permalink]

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30 Sep 2007, 00:54
singh_amit19 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

h=-16(t-3)^2+150
the object reaches its maximum height at t=3
then back down. after two seconds i guess it is 134
C
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30 Sep 2007, 02:09
h = -16 (t - 3)^2 + 150.

just plug values for t !

t = 1 then h = 86
t = 2 then h = 134
t = 3 then h = 150
t = 4 then h = 134
t = 5 then h = 86

As you can see, the maximum height is h = 150 at t=3 so after two seconds the height will be 86.

No junior high physics is needed.

Senior Manager
Joined: 11 Sep 2005
Posts: 324
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30 Sep 2007, 04:06
KillerSquirrel wrote:
h = -16 (t - 3)^2 + 150.

just plug values for t !

t = 1 then h = 86
t = 2 then h = 134
t = 3 then h = 150
t = 4 then h = 134
t = 5 then h = 86

As you can see, the maximum height is h = 150 at t=3 so after two seconds the height will be 86.

No junior high physics is needed.

BINGO! Now, that's wht i call a real Ivy league stuff............good one KillerSquirrel
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30 Sep 2007, 12:14
KillerSquirrel wrote:
h = -16 (t - 3)^2 + 150.

just plug values for t !

t = 1 then h = 86
t = 2 then h = 134
t = 3 then h = 150
t = 4 then h = 134
t = 5 then h = 86

As you can see, the maximum height is h = 150 at t=3 so after two seconds the height will be 86.

No junior high physics is needed.

lol, i tried to apply newton's gravity findings
30 Sep 2007, 12:14
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