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Mathematics 8 Online
OpenStudy (epicteatime):

How long will the water stay in the air? y=-16x^2+233x+0

OpenStudy (epicteatime):

@aaronq

OpenStudy (epicteatime):

@iPwnBunnies

OpenStudy (epicteatime):

H(t)= -16t^2+vt+s

OpenStudy (epicteatime):

a=-16, b=233,c=0

OpenStudy (ipwnbunnies):

Ok, this is a position function, given. You're trying to find the time, t, when the water hits the ground. Do you agree when the function equals 0, the water is at ground level?

OpenStudy (ipwnbunnies):

So, the function is y = -16t^2 + 233x?

OpenStudy (epicteatime):

yeah, thats why i put 0 there silly ;3

OpenStudy (ipwnbunnies):

233t*

OpenStudy (epicteatime):

anyway...

OpenStudy (epicteatime):

lets get to les solving!

OpenStudy (ipwnbunnies):

Nooo, you added a 0. Which is not the same as setting the equation equal to 0.

OpenStudy (epicteatime):

0 DOESNT HAVE TO BE THERE!!

OpenStudy (epicteatime):

0=-16t^2+233x

OpenStudy (epicteatime):

there is that better for you?

OpenStudy (ipwnbunnies):

Yesh. 0 = -16t^2 + 233t 16t^2 = 233t Reduce, solve for t. That'll be the time when the water reaches the ground, when the position function equals 0.

OpenStudy (epicteatime):

whaat?

OpenStudy (epicteatime):

i leant this a totally different way ;P

OpenStudy (ipwnbunnies):

OHHHH. I read your question wrong, I think. How long will it stay in the air. Smh.

OpenStudy (epicteatime):

yush

OpenStudy (epicteatime):

you must use the other equation, with the -b and the \[\pm\]

OpenStudy (epicteatime):

and the

OpenStudy (epicteatime):

thing

OpenStudy (ipwnbunnies):

Quadratic formula? Idts. If I remember correctly... When t = 0, initial point, the position function equals 0, y(0) = -16(0)^2 + 233(0) = 0 I advise you to find the vertex of this function. The x-value will be the time it takes the water to reach its maximum height. Multiply that time by 2 to get the total time the water will be in the air.

OpenStudy (epicteatime):

yeah that ;P

OpenStudy (epicteatime):

-b/2a

OpenStudy (epicteatime):

7.28125 ;3

OpenStudy (epicteatime):

is the vertex

OpenStudy (ipwnbunnies):

Right, is the x-value of the vertex, which is the same as the time it'll take to get there. Then multiply by 2, you'll get the total time. :D

OpenStudy (epicteatime):

uhhh ok ;3

OpenStudy (epicteatime):

14.5625?

OpenStudy (epicteatime):

but but....

OpenStudy (ipwnbunnies):

Yes. :) Use a graphing utility to verify the vertex is indeed at that point. :3

OpenStudy (epicteatime):

yeah

OpenStudy (epicteatime):

so what shall i do...hmmmmmmm

OpenStudy (ipwnbunnies):

??

OpenStudy (epicteatime):

?????

OpenStudy (epicteatime):

i have no idea of any 'graphing technology' which i can use ;P

OpenStudy (ipwnbunnies):

What shall you do? Lol

OpenStudy (ipwnbunnies):

Ohh, it's ok. I just verified it. xD

OpenStudy (epicteatime):

oh ok

OpenStudy (epicteatime):

i need to find the x-axis's of the thing ;P which i forgot how to do, other than the tiresome way using the table ;3

OpenStudy (ipwnbunnies):

I don't think I understand what you're saying lolol

OpenStudy (epicteatime):

Use the graphing technology of your choice to graph the above equation and determine the x-intercept values. Set rounding to 5 places. Save a copy of this graph and paste or attach it to this Worksheet.

OpenStudy (epicteatime):

there ;3

OpenStudy (epicteatime):

.............

OpenStudy (ipwnbunnies):

Omgg. You know what. The x-intercepts would actually give you the same answer. .-. I'm a knucklehead. Sorry for making this more complicated. :3

OpenStudy (epicteatime):

uhhhhhhu ;P

OpenStudy (epicteatime):

i stilly dunno

OpenStudy (epicteatime):

.....

OpenStudy (epicteatime):

@aaronq

OpenStudy (epicteatime):

oh you're back ;3

OpenStudy (ipwnbunnies):

Oh, using a graphing calculator? Ehh. I don't think I'm good at explaining that at all, lol. I'm not too great at using the graphing utilities myself.

OpenStudy (epicteatime):

lol

OpenStudy (epicteatime):

@aaronq

OpenStudy (epicteatime):

@Hero

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