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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jainrahul1985
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user123321
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if you observe the equation h=-16(t-3)^2 + 150jainrahul1985 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
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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GMATGuruNY
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To MAXIMIZE h = -16(t - 3)² + 150, we need to MINIMIZE 16(t-3)².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
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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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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