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

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Manager
Joined: 03 Jul 2008
Posts: 62

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

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05 Jul 2008, 11:37
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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Joined: 17 Nov 2007
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Re: An object thrown directly ..... [#permalink]

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05 Jul 2008, 12:04
1
KUDOS
Expert's post
B

1. h=hmax at t=3
2. t1=3+2=5 -->h1= -16*4+150=86
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Kudos [?]: 4478 [1], given: 360

Manager
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Posts: 62

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Re: An object thrown directly ..... [#permalink]

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05 Jul 2008, 12:22
walker wrote:
B

1. h=hmax at t=3
2. t1=3+2=5 -->h1= -16*4+150=86

hi Walker I don't understand your explanation. how did you know height is at its maximum when t = 3, where did you get the 4 from in part 2. thanks
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Re: An object thrown directly ..... [#permalink]

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05 Jul 2008, 13:25
1
KUDOS
TheBigCheese 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 will be max when t = 3 because in the eq: "h = -16 (t-3)^2 + 150", "-16 (t-3)^2" is -ve. we need to make this part to minimum to get the max hight. when is "h" max? when t = 0. so, 2 second after it reaches its max height in 3 sec, it will be at:
h = -16 (5-3)^2 + 150 = 86

why t = 5? because it takes 5 sec to arrive at 86 after it reaches to its max. height.
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Re: An object thrown directly ..... [#permalink]

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05 Jul 2008, 13:52
GMAT TIGER wrote:
TheBigCheese 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 will be max when t = 3 because in the eq: "h = -16 (t-3)^2 + 150", "-16 (t-3)^2" is -ve. we need to make this part to minimum to get the max hight. when is "h" max? when t = 0. so, 2 second after it reaches its max height in 3 sec, it will be at:
h = -16 (5-3)^2 + 150 = 86

why t = 5? because it takes 5 sec to arrive at 86 after it reaches to its max. height.

thanks kudos to you both
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Re: An object thrown directly ..... [#permalink]

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06 Jul 2008, 18:31
TheBigCheese 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

To know the height,2 seconds after ithe object reaches the maximum height we need to know the time required to reach the max height.
h = -16 (t-3)^2 + 150
h = 150 -16 (t-3)^2
hmax the bolded term must be minimum
=> t= 3

total time = 2+3 =5
subst this for t in h we h = 86

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Re: An object thrown directly .....   [#permalink] 06 Jul 2008, 18:31
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