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Thread: Height of a ball...

20130513, 01:37 #1
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Height of a ball...
An SAT question.
h(t) = c  (d  4t)^2
At a time t = 0, a ball was thrown upward from an initial height of 6 feet. Until the ball hit the ground, its height, in feet, after t seconds was given by the function h above, in which c and d are positive constants. If the ball reached its maximum height of 106 feet at a time t = 2.5, what was the height, in feet, of the ball at time t = 1?

20130513, 08:00 #2
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Gosh, doing these was a long time ago! How about
BATcher
Time prevents everything happening all at once...

20130513, 11:13 #3
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I don't know how to do the spoiler thing... so I will write out my answer with words and make it tiny and really light instead, ok? if you want to read it, highlight my post, it's much easier to read
fourtytwo point four feet
I'm no math major, but isn't the formulae a misnomer in this question? You never asked what 'c' or 'd' was, so who cares about that side of the equation? and the initial 6ft of height is included in the 106ft... so you can ignore it as another red herring...? if my logic is flawed, please discuss.
so you are asking basically;
106(2.5)
??(2.0)
??(1.5)
??(1.0) < that one
if 2.5 seconds pass after the ball has been released and the ball is currently at it's peak height of 106ft, how high was it after only a single second passed after the ball was released?
No initial velocity.
I get the impression I'm supposed to use gravity somehow.... but for that we need the weight of the ball to begin with, right? or am I supposed to somehow use that equation to deduce all this other information?
so I approach this problem as a 'factoring' problem (term?)
if 106(h) = 2.5(t)
and t now equals 1
what does h now equal?
Last edited by Ben09880; 20130513 at 11:21.

20130513, 11:58 #4
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BATcher, you are correct. (I don't know how to do the spoiler thing either...where are those instructions?  that's a puzzle in itself).
Ben, the problem was written correctly from the SAT sample test.

20130513, 12:04 #5
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well, guess that explains why I couldn't get through college algebra II
so you *do* need to use the *whole thing*.
dang it.
I'm getting too old for this... :P hahaha

20130513, 12:09 #6
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[ spoiler]what you want to hide[ /spoiler]
Remove the obvious blank after each leftsquarebracket!BATcher
Time prevents everything happening all at once...

20130513, 12:17 #7
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From: h(t) = c  (d  4t)^2
Start with t=0. Then, h(0) = c  d^2 = 6.
Then, h(2.5) = 106 = c  (d  10)^2 and 106 = c  d^2 + 20d  100 and 206 = 6 + 20d and 200 = 20d, so d = 10
That means c = 106
Then, h(1) = c  (d  4)^2 = c  d^2 + 8d  16 = 6 + 80  16 = 70 (the solution)

20130514, 01:25 #8
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You forgot the "QED" !
BATcher
Time prevents everything happening all at once...