Results 1 to 8 of 8
Thread: Height of a ball...

20130513, 00:37 #1
 Join Date
 Jan 2001
 Location
 La Jolla, CA
 Posts
 1,509
 Thanks
 35
 Thanked 66 Times in 62 Posts
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, 07:00 #2
 Join Date
 Feb 2008
 Location
 A cultural area in SW England
 Posts
 3,472
 Thanks
 34
 Thanked 197 Times in 177 Posts
Gosh, doing these was a long time ago! How about
BATcher
"The trouble with quotes on the internet is that you can never know if they are genuine."
Abraham Lincoln

20130513, 10:13 #3
 Join Date
 Apr 2013
 Posts
 103
 Thanks
 9
 Thanked 10 Times in 9 Posts
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 10:21.

20130513, 10:58 #4
 Join Date
 Jan 2001
 Location
 La Jolla, CA
 Posts
 1,509
 Thanks
 35
 Thanked 66 Times in 62 Posts
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, 11:04 #5
 Join Date
 Apr 2013
 Posts
 103
 Thanks
 9
 Thanked 10 Times in 9 Posts
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, 11:09 #6
 Join Date
 Feb 2008
 Location
 A cultural area in SW England
 Posts
 3,472
 Thanks
 34
 Thanked 197 Times in 177 Posts
[ spoiler]what you want to hide[ /spoiler]
Remove the obvious blank after each leftsquarebracket!BATcher
"The trouble with quotes on the internet is that you can never know if they are genuine."
Abraham Lincoln

20130513, 11:17 #7
 Join Date
 Jan 2001
 Location
 La Jolla, CA
 Posts
 1,509
 Thanks
 35
 Thanked 66 Times in 62 Posts
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, 00:25 #8
 Join Date
 Feb 2008
 Location
 A cultural area in SW England
 Posts
 3,472
 Thanks
 34
 Thanked 197 Times in 177 Posts
You forgot the "QED" !
BATcher
"The trouble with quotes on the internet is that you can never know if they are genuine."
Abraham Lincoln