<p>h(t) = c - (d - 4t)^2</p>
<p>At time t=0, a blal was thrown upward from an initial height of 6 feet. Until the ball hit the ground, its height, in feet, afer t seconds was given by the function h above, in which c and d are positive constants. If the ball reached its maximum hiehgt of 106 feet at time t=2.5, what was the hieght, in feet, of the ball at time t=1?</p>
<p>Blue book p. 657, problem 18.</p>
<p>Never mind, I used the consolidated solutions. My bad.</p>
<p>Can someone provide an explination of how to do this problem? Cant seem to get it.</p>
<p>h = c - (d-4t)^2
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>I made my math teacher do it for me. she's an extremely bright latley and had no idea what to do at first. we ended up squaring it and then cancelling and stuff. that was a really rediculous question though.</p>
<p>the science nerd inside of me tried to solve it the physics way, but that didn't work haha</p>
<p>Ok i see how to do it. Yeah i guess they do give you the height at 0 for a reason lol.</p>
<p>In short.</p>
<p>h(t) = c - (d-4t)^2
h(t) will be at its maximum=c when d-4t=0, t=d/4.
The parabola vertex is in (d/4, c).
(You can also reach this result based on the standard form of the equation of a parabola.)
Since maximum height of the ball is 106 feet at time t=2.5, the vertex is in (2.5, 106).
d/4=2.5, d=10, and
c=106.
Plugging this values and t=1 into
h(t) = c - (d - 4t )^2,
h(1) = 106 - (10 - 4)^2 = 70 feet at time t=1.</p>
<p>Why did they give us the height at time t=0? To lead us on a wilde-goose chase? Evil! :D</p>
<p>if at time 0 you are at 6, doesn't the ball only travel 100 after 2.5 seconds?</p>
<p>Yes it does.</p>
<p>so then why do u use 106 when it only travels 100..</p>
<p>The formula gives us the absolute height - not the change of height from time 0.
At t=2.5 the height is 106.</p>