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height of h feet after t

by maihuna » Sat Dec 12, 2009 11:39 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
Answer:
Last edited by maihuna on Sat Dec 12, 2009 1:00 pm, edited 1 time in total.
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by Stuart@KaplanGMAT » Sat Dec 12, 2009 12:48 pm
maihuna 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
Answer:
Let's think about our equation:

h = -16 (t - 3)2 + 150

(I'm assuming that the 2 after the brackets means that we really have (t-3)^2, not (t-3)*2.)

Since (t-3)^2 will always be non-negative, and since -16 is negative, the first term, -16*(t-3)^2, will always be non-positive.

So, in order to maximize height, we want that first term to be 0, which will occur when (t-3)^2 = 0, which will occur when t=3.

Therefore, we reach our max height of 150 when t=3.

We want to know our height 2 seconds after that, so let's let t=5 and solve!

h = -16*(t-3)^2 + 150

h = -16*(5-3)^2 + 150

h = -16*(2)^2 + 150

h = -16*4 + 150

h = -64 + 150

h = 86... choose B.
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by maihuna » Sat Dec 12, 2009 12:59 pm
maihuna 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
Answer:
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Legendary Member
Posts: 1578
Joined: Sun Dec 28, 2008 1:49 am
Thanked: 82 times
Followed by:9 members
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by maihuna » Sat Dec 12, 2009 1:02 pm
Thanks Stuart yes it is ^2
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