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# Height of an object thrown upwards

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Manager
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Height of an object thrown upwards [#permalink]

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26 Nov 2007, 14:44
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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
_________________

--gregspirited

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Re: Height of an object thrown upwards [#permalink]

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26 Nov 2007, 15:07
gregspirited 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

hmm interesting:

got B. will write more later if it ic correct.
CEO
Joined: 29 Mar 2007
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Re: Height of an object thrown upwards [#permalink]

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26 Nov 2007, 17:43
gregspirited 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

Is this written correctly? why do we have -16(t-3)2? why not just

-32(t-3). Just making sure i am reading this correctly.
Director
Joined: 22 Aug 2007
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Re: Height of an object thrown upwards [#permalink]

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26 Nov 2007, 17:57
gregspirited 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

B,

The object reaches its max height when t=3.
To find the height two seconds after the object reaches its max height means to find height when t=5.

just plug in 5
Manager
Affiliations: CFA L3 Candidate, Grad w/ Highest Honors
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Posts: 130
Location: USA
Schools: Chicago Booth R2 (WL), Wharton R2 w/ int, Kellogg R2 w/ int
WE 1: Global Operations (Futures & Portfolio Financing) - Hedge Fund ($10bn+ Multi-Strat) WE 2: Investment Analyst (Credit strategies) - Fund of Hedge Fund ($10bn+ Multi-Strat)
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Re: Height of an object thrown upwards [#permalink]

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26 Nov 2007, 18:41
GMATBLACKBELT wrote:
gregspirited 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

Is this written correctly? why do we have -16(t-3)2? why not just

-32(t-3). Just making sure i am reading this correctly.

Looks like a bogus question to me. I dont see how it could be written correctly. If "h = -16 (t - 3)2 + 150." is right then it reaches maximum height at ZERO seconds. h = -32(0-3) + 150 = 236
Director
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Re: Height of an object thrown upwards [#permalink]

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26 Nov 2007, 18:58
robertrdzak wrote:
GMATBLACKBELT wrote:
gregspirited 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

Is this written correctly? why do we have -16(t-3)2? why not just

-32(t-3). Just making sure i am reading this correctly.

Looks like a bogus question to me. I dont see how it could be written correctly. If "h = -16 (t - 3)2 + 150." is right then it reaches maximum height at ZERO seconds. h = -32(0-3) + 150 = 236

t can not be negative, and it can not be zero, zero would mean the object did not move at all, and if it is thrown upwards, then h (t=0) is defenetely is not the max height.

but it makes sense that the closer t is to zero the higher is h, let`s say t is very close to zero, then h after 2 seconds must be very close to h (t=2)

h=-16(2-3)2+150=182....hmm
Senior Manager
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Re: Height of an object thrown upwards [#permalink]

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26 Nov 2007, 19:13
gregspirited 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

I thought when t = 1 - you get the max height.

-16(-2)2 + 150 = 214

and the fall in height in 2 sec is

-16(-1)2 + 150 = 182.

the diff. I got was 32 ..but none of the answer choices is 32. Dont know where I am going wrong.
Manager
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Height of an object thrown upwards ... Wrong Equation [#permalink]

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26 Nov 2007, 19:56
The equation is incorrect to start with. As I querried my system for past discussions on this particular question, I found that the original equation is:

h=-16(t-3)^2+150

I think that explains why gregspirited had the confusion, triggering all the follow-up posts. Also, this question was debated for a long time on beatthegmat.com forums. I'm including a link for the discussion thread on beatthegmat.com to help those curious souls that have enough patience and interest in seeing what all possible solutions were arrived at when the posted the wrong equation..

http://www.beatthegmat.com/set-26-q-21-t5149.html#20127 For those having difficulty accessing the URL, I've made a PDF file out of the discussions on Beatthegmat.com and attached it here..Hope this helps. Good luck.
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File comment: Height of an object thrown upwards...
set-26-q-21-t5149.pdf [109.51 KiB]

Manager
Joined: 07 Oct 2005
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27 Nov 2007, 14:01
Sorry Guys...My equation was wrong as wizard had posted. It should be ^2 and not *2. I was in bit hurry when I posted.
But when ^2 it is a valid question and good one too.

_________________

--gregspirited

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