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jainrahul1985: Master | Next Rank: 500 Posts; Posts: 228; Joined: Sun Aug 17, 2008 8:08 am; Thanked: 4 times

by jainrahul1985 » Fri Dec 30, 2011 9:19 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

OA B

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Source: Beat The GMAT — Problem Solving | Fri Dec 30, 2011 9:19 pm

user123321: Master | Next Rank: 500 Posts; Posts: 385; Joined: Fri Sep 23, 2011 9:02 pmFollowed by: 6 members; Thanked: 62 times

by user123321 » Fri Dec 30, 2011 9:33 pm

jainrahul1985 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

OA B

if you observe the equation h=-16(t-3)^2 + 150
h will have max value at t=3(and also t>=0 as it is time)
anything other than that will actually reduce h.
so it reaches max height after 3 seconds.

height after 5 sec = -16(2)^2 + 150
= 86

user123321

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Want to do it right the first time.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sat Dec 31, 2011 8:33 am

jainrahul1985 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

OA B

To MAXIMIZE h = -16(t - 3)Â² + 150, we need to MINIMIZE 16(t-3)Â².
Since (t-3)Â² cannot be negative, the smallest possible value of 16(t-3)Â² is 0.
16(t-3)Â² = 0 when t=3.
Two seconds later, t=5.
When t=5, h = -16(5-3)Â² + 150 = -64 + 150 = 86.

The correct answer is B.

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