<p>pg 657 from the CB book number 18. I dont know how to get the answer that they have, can someone explain?</p>
<p>This is the most tedious problem in the book. Sigh, I highly doubt we would get something like this on the real test.</p>
<p><a href="http://talk.collegeconfidential.com/showpost.php?p=1972318&postcount=4%5B/url%5D">http://talk.collegeconfidential.com/showpost.php?p=1972318&postcount=4</a>
Credited to obsessedAndre</p>
<p>We are given the height is 6 when t = 0, therefore
6 = c-(d-0*t)^2</p>
<p>6 = c - d^2</p>
<p>at t = 2.5, the height is 106</p>
<p>106 = c - (d - 4*2.5)^2
106 = c - (d - 10)^2</p>
<p>expand the binomial, being very careful to distribute the negative sign</p>
<p>106 = c - d^2 + 20d - 100
106 = (6) + 20d - 100
200 = 20d
d = 10</p>
<p>now go back to c-d^2 = 6 and solve for c
c - 10^2 = 6
c - 100 = 6
c = 106</p>
<p>now it's easy</p>
<p>h = 106 - (10 - 4)^2
h = 106 - 36
h = 70</p>
<p>oh ok now i understand, but that is a long method and takes a while to do. i hope you are right and we dont get a problem like this.</p>
<p>since it appeared in a CB test, i think it's possible that you might get a question like that. i think it's extremely unlikely, but possible. so i was wondering if anybody has a faster way to solve it? I've been thinking about this for a couple days and i can't come up with one, but i strongly suspect that it exists.</p>
<p>as a step in that direction, i feel like it has to be telling that 106 is 6 + 10^2, and 70 is 6 + 8^2. does that help in any way?</p>
<p>There are several discussions of this question (657/18) in
Consolidated</a> List of Blue Book Math Solutions, 3rd ed.</p>
<p>This approach seems a bit shorter:
<a href="http://talk.collegeconfidential.com/showthread.php?p=4057931#post4057931%5B/url%5D">http://talk.collegeconfidential.com/showthread.php?p=4057931#post4057931</a>.</p>
<p>The last line should read
h(1) = 106 - (10 4(1))^2 = 106 6^2 = 70 feet.</p>
<p>How does CB propose solving this?</p>
<p>At 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 below, in which c and d are positive constants. If the ball reached its maximum height of 106 feet at time t=2.5, what was the height, in feet, of the ball at time t=1?</p>
<p>h(t) = c - (d - 4t)^2</p>
<p>SOLUTION: We know (0,6) lies on the curve. We also know (2.5, 106) is on the curve and is the vertex. </p>
<p>Expand the function:
c - (d^2 -8dt + 16t^2) = h(t)
c - d^2 +8dt -16t^2 = h(t)
Plugging in the first point we have:
c - d^2 = 6
Plugging in the second point we have:
c - d^2 + 8d(5/2) - 16(5/2)^2 = 106
c - d^2 + 20d - 100 = 106
Plug in 6 for c-d^2 in the second equation:
6 + 20d - 100 = 106
20d = 200
d=10
Plug d=10 into c-d^2 = 6:
c - 100 = 6
c = 106</p>
<p>So we have:
h(t) = 106 - (10-4t)^2
Plug in 1 for t.
h(1) = 106 - (10-4)^2
h(1) = 106 - 36
h(1) = 70</p>
<p>Solved.</p>