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
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sudhir3127
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The object reaches maximum height at t=3 seconds.
At t=3+2=5 seconds ( 2 seconds later)
h = [-16*(5-3)*2] + 150 = [-16*4] +150 = 86 feet.
Answer is B.
At t=3+2=5 seconds ( 2 seconds later)
h = [-16*(5-3)*2] + 150 = [-16*4] +150 = 86 feet.
Answer is B.
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santa_dem
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I didn't get the 3 seconds either.
Yes, when t=3 h=150, but what if t=1, then h= -16*(-2)*2+150=216
or, if t=2, h=182.
I would agree to "h is max when t=3" if it were |t-3| instead of (t-3), or (t-3)^2 instead of (t-3)2.
Could someone please explain.
Thanks.
Yes, when t=3 h=150, but what if t=1, then h= -16*(-2)*2+150=216
or, if t=2, h=182.
I would agree to "h is max when t=3" if it were |t-3| instead of (t-3), or (t-3)^2 instead of (t-3)2.
Could someone please explain.
Thanks.
Last edited by santa_dem on Wed Aug 13, 2008 7:36 am, edited 1 time in total.
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anksbhandari
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Ian Stewart
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The equation is surely supposed to read:CITI29 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
h = -16 (t - 3)^2 + 150
i.e. the 2 should be an exponent. If you rewrite it as
h = 150 - 16(t-3)^2
you can see that we want (t-3)^2 to be as small as possible to make h as large as possible- that is, we want t to be equal to 3 for maximum height. Then to answer the question, we just need to plug in t = 5.
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