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by CITI29 » Tue Aug 12, 2008 1:56 pm
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 sudhir3127 » Tue Aug 12, 2008 8:52 pm
The object reaches maximum height at t=3 seconds.

At t=3+2=5 seconds ( 2 seconds later)

h = [-16*(5-3)*2] + 150 = [-16*4] +150 = 86 feet.


Answer is B.
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by CITI29 » Wed Aug 13, 2008 6:26 am
How do u know object reaches its max height in 3 seconds?
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by santa_dem » Wed Aug 13, 2008 7:11 am
I didn't get the 3 seconds either.

Yes, when t=3 h=150, but what if t=1, then h= -16*(-2)*2+150=216

or, if t=2, h=182.


I would agree to "h is max when t=3" if it were |t-3| instead of (t-3), or (t-3)^2 instead of (t-3)2.

Could someone please explain.

Thanks.
Last edited by santa_dem on Wed Aug 13, 2008 7:36 am, edited 1 time in total.
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by anksbhandari » Wed Aug 13, 2008 7:19 am
i think this question is wrongly framed
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Re: height

by Ian Stewart » Wed Aug 13, 2008 8:47 am
CITI29 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
The equation is surely supposed to read:

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

i.e. the 2 should be an exponent. If you rewrite it as

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

you can see that we want (t-3)^2 to be as small as possible to make h as large as possible- that is, we want t to be equal to 3 for maximum height. Then to answer the question, we just need to plug in t = 5.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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