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An object thrown directly upward is at a height of h feet af
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24 Jul 2020, 07:12
24 Jul 2020, 07:12
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Question Stats:
60% (01:27) correct
40% (01:24) wrong based on 40 sessions
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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
A. 6
B. 86
C. 134
D. 150
E. 214
Re: An object thrown directly upward is at a height of h feet af
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24 Jul 2020, 07:13
24 Jul 2020, 07:13
Expert Reply
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Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
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Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
For more GRE Quant Theory and Practice, see
GRE Prep Club - MATH Book for the GRE Exam
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Re: An object thrown directly upward is at a height of h feet af
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24 Jul 2020, 07:33
24 Jul 2020, 07:33
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Carcass 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
A. 6
B. 86
C. 134
D. 150
E. 214
The formula h = -16 (t - 3)² + 150 allows us to determine the height of the object at any time. For what value of t is -16(t-3)² + 150 MAXIMIZED(in other words, the object is at its maximum height)?
It might be easier to answer this question if we rewrite the formula as h = 150 - 16(t-3)²
To MAXIMIZE the value of h, we need to MINIMIZE the value of 16(t-3)² and this means minimizing the value of (t-3)²
As you can see,(t-3)² is minimized when t = 3.
We want to know the height 2 seconds AFTER the object's height is maximized, so we want to know that height at 5 seconds (3+2)
At t = 5, the height = 150 - 16(5 - 3)²
= 150 - 16(2)²
= 150 - 64
= 86
Answer: B
