Math question..

Test Preparation
1

<p>Came across this taking a practice test. It was the last Math question in the section, and was a grid-in, and I have not a clue how to solve it.</p>

<p>h(t) = c - (d - 4t)²</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 above, 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>Any clue how to go about this?</p>

2

<p>I believe you can figure out the the value of c and d using the coordinate (2.5,106) and the fact that the equation seems to be in vertex form.</p>

3

<p>The problem can be solved using muti step substitution.</p>

<p>h(t) = c -(d-4t)^2
When t = 0 h = 6 so</p>

<p>6 = c -(d-0)^2
6 = c -d^2</p>

<p>When t= 2.5 h = 106</p>

<p>106 = c - (d-10)2
106 = c - d^2 +20d -100
206 = c - d^2 + 20d
Remember c-d^2 = 6
206 = 6 + 20D
d = 10</p>

<p>106 = c - (10-10)^2
106 = c</p>

<p>h(t) = 106 - (10-4t)^2
h(1) = 106 - (10-4)^2
h(1) = 106 - 36
h(1) = 70</p>

4

<p>Ahh, thanks so much!</p>

<p>I got up to where you substitute 6 for c - d^2, but messed up with the signs.. I forgot to switch them, so I was getting 106 = 6 + 20d + 100, and then just getting d = 0.</p>

5

<p>It makes more sense to ask these questions and look for answers in "SAT Preparation" forum.
" Updated Consolidated List of Blue Book Math Solutions" thread has a solution to your question:
657 / 18 /</p>