OG 13th Edition Question 129, PS

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OG 13th Edition Question 129, PS

by babakp » Tue Mar 26, 2013 12:07 am
Hi everybody,

I am new and I hope it is ok to post this question here.

Could please someone be so kind and help in with this question.

I don't really understand the answer from the book.

Many thanks. Community Manager
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by neelgandham » Tue Mar 26, 2013 1:10 am
For the benefit of other users, here is the question:
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)166
h = -16 (t - 3)^2 + 150.
h = ("Negative number" * "Square of a number") + 150.
Maximum value of h is when the "Square of a number" is 0.
h = -16 (t - 3)^2 + 150.
h is maximum when t = 3 . For any t greater than or less than 3, the value of h is less than 150.

In conclusion, when t = 3 secs, the object is at the maximum height of 150 meters.
2 seconds after it reached the maximum height, t = 5

h = -16 (5 - 3)^2 + 150 feet.
h = (-16 * 4) + 150 feet.
h = -64 + 150 feet.
h = 86 feet.

Let me know if still need help.

p.s: It is so good to be back!
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by GMATGuruNY » Tue Mar 26, 2013 2:09 am
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
Equation, rephrased:
h = 150 - 16t - 3)Â².

To MAXIMIZE the value of h, we need to MINIMIZE the value SUBTRACTED from 150: 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.
Thus, the maximum height occurs when t=3.

Two seconds later, t=5.
When t=5, h = 150 - 16(5-3)Â² = 150-64 = 86.

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by [email protected] » Tue Mar 26, 2013 6:01 am
aman88 wrote:Q: 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
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

Cheers,
Brent

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by babakp » Tue Mar 26, 2013 11:31 am
All, thank you very much.

It is now totally clear to me!

Glad to be on this forum.

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