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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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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

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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)² + 150, we need to MINIMIZE 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.
Two seconds later, t=5.
When t=5, h = -16(5-3)² + 150 = -64 + 150 = 86.

The correct answer is B.
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