This question is tough for me.. Help!

[yumi2012]

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

[GMATGuruNY]

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

Equation, rephrased:
h = 150 - 16(t-3)².

To MAXIMIZE the value of h, we need to MINIMIZE the value SUBTRACTED from 150: 16(t-3)².
Since (t-3)² cannot be negative, the smallest possible value of 16(t-3)² is 0.
16(t-3)² = 0 when t=3.
Thus, the maximum height occurs when t=3.

Two seconds later, t=5.
When t=5, h = 150 - 16(5-3)² = 150-64 = 86.

The correct answer is B.

[faraz_jeddah]

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

Equation, rephrased:
h = 150 - 16t - 3)².

To MAXIMIZE the value of h, we need to MINIMIZE the value SUBTRACTED from 150: 16(t-3)².
Since (t-3)² cannot be negative, the smallest possible value of 16(t-3)² is 0.
16(t-3)² = 0 when t=3.
Thus, the maximum height occurs when t=3.

Two seconds later, t=5.
When t=5, h = 150 - 16(5-3)² = 150-64 = 86.

The correct answer is B.

This is a good question. What is the source?

Also, Mitch - Do we need to use some physics and understand that the object is on its way down after reaching its max. height?

[[email protected]]

Hi faraz_jeddah,

Yes, logically you do have to realize that since the ball is thrown upward, once it hits its maximum height, then it's heading back down. The essential part is figuring out what the maximum height is; once that's taken care of, then you can just +2 to the time and solve for height.

GMAT assassins aren't born, they're made,
Rich

[macattack]

This is a typical Max/Min problem. Usually in max min problems you have an equation of the form: Negative*variable+constant. Now to maximize that equation you have to minimize the negative part of it!