Halp... Algebra II >_<

[Aletheia]

Is anybody here good at quadratic functions? I have to do a project for my class where I toss a basketball up into the air three times, and each time measure the time it took to hit the ground and the distance above the ground from which I threw it. I have all my measurements, and I'm trying to do the calculations, but it doesn't feel like it's coming out right.

Measurements:

1st Toss: time - 1.8 seconds, distance - 3'4"
2nd Toss: time - 1.8 seconds, distance - 3'4"
3rd Toss: time - 1.7 seconds, distance - 3'4"

The form I'm supposed to use for the function is h(t) = -16t² + vt + s, t = time, v = velocity, s = distance from the ground. I have to solve for v first, then find the vertex of the parabola by completing the square and writing the function in vertex form.

I did all this, and then tried graphing it on my calculator. The standard form looks about right, but when I graph the vertex form I get something way different. Is that supposed to happen? Also, the vertex I get from putting it in vertex form looks way off.

I hope this makes sense. Halp pleez. ^_^

[ADXC]

Well, I try to help you.

1st toss
-16t^2 + vt +s First plug in your information. Im gonna skip a step or two to save time.
-51.84 + v(1.8) + 40 I changed the form of 3'4" to 40 inches.
11.84= 1.8V
11.84/1.8=1.8v/1.8
v=6.57 Okay so Im not really sure about this answer
-16t^2 + 6.57t +s Complete the square.
(-16t^2 + 6.57t +10.816) + 40 -10.816
" " +29.184
I seriously doubt that I could factor this so Im just going to use the quadratic formula.
x= -6.57 +/- Square root of 43.1649 - 4(-16)(10.816) all / 2(-16)
x= -6.57 +/- Square roof of 735.3889 all/ -32

Well Im really sorry, I forget what to do next. Im not even sure Im doing this right. Im sorry to have wasted your time, please forgive me!

[Aletheia]

Hmm... that's pretty much what I did, except that I used 3.33 ft instead of 40 in, and I factored out -16 before completing the square. I think I'm right... but oh well, I already printed it all out and glued it onto the poster, and the project is due tomorrow, so there's nothing I can really do about it now. Thanks for the help, though. ^^

[Icarus]

This is what I get for not being online.


Just going from ADXC's,

Complete the square:
-16t^2 + 6.57t + s =0

-16t^2 + 6.57t + 3.33 = 0 = 16*t^2 - 6.57*t - 3.33
// the distance does need to be in feet, as the formula she gave you has the acceleration
// from gravity in fps squared. Also, since turning the problem on it's head doesn't affect
// the zeros, I'll make the leading term positive.

16*t^2 - 6.57*t + 3.285^2= 3.333 + 3.285^2
//when completing the square, you take the square root of the first term (16t^2) and the
//last term on the left side, and of the sum on the right.

4*t - 3.285 = sqrt(14.125)
// we can ignore the negative root, as negative time makes no sense.

4t- 3.285 = 3.758

4t= 7.043

t= 1.76 // which is roughly what you had.

If I'm not mistaken, the vertex formula says that the ball will reach its peak at time t such that

32*t= 6.57 (from 2*a*t +b).

Graphing that will yield a different picture than the original equation, and a time of about a fifth of a second.

Max height (plugging .2 in for t) seems to be about 4 feet.

Hope it helps.

[ADXC]

Ugh, yeah I sorta lost some of my Algebra 2 knowledge since last year. Im now taking precal which is well more difficult. Im just about tired of logarithms.

[Aletheia]

Well, thanks guys. I already turned in my project, so if it's wrong, it's wrong. But hopefully this will help in the future. ^^