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Created: 2011/12/31, Members: 42, Messages: 22740

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Algebra Problem

Anni313
Anni313
Posts: 1,790
Joined: 2004/03/04
United States
2005/02/14, 08:57 PM
Okay, I totally give up.

A person throws a rock upward from the edge of an 80-foot cliff. The height, H, in feet, of the rock above the water at the bottom of the cliff after x seconds is described by the function.

H(x)=-16(x squared) + 64x + 80

a)what is the height of the rock after 3 seconds?

I came up with 128 feet

b)After how many seconds will the rock be at it's highest, and what will that height be?

c)How long will it take the rock to reach the water. Hint the height would be 0) You may use the quadratic formula or the factoring technique.

Well that's just hunky dory. Why are we throwin rocks anyway? What if we put somebody's eye out? Sigh....

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Anni

*******
In my head, I am out of my mind....
Pritchard
Pritchard
Posts: 1,212
Joined: 2004/03/02
United Kingdom
2005/02/15, 06:04 AM
i dont know |:

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wax on, wax off
asimmer
asimmer
Posts: 8,201
Joined: 2003/01/07
United States
2005/02/15, 07:28 AM
And how does this apply to anything in real life? i don't get why we want to know anything other than where not to stand when the rock comes back down...

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Vision without action is a daydream, Action without vision is a nightmare.
-- Japanese Proverb --

Pritchard
Pritchard
Posts: 1,212
Joined: 2004/03/02
United Kingdom
2005/02/15, 07:31 AM
anyway, people in glass houses shouldnt throw stones!
(i have no idea how it relates the problem, but i wanted to use it!)

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wax on, wax off
jonathanweaver
jonathanweaver
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United States
2005/02/15, 11:16 AM
Using the quadratic formula, I got that when H(x)=0, x=5 or -1. Since we are talking about time, I'd have to go with the 5. However, I'm questioning my answer based on the fact that after 3 seconds it was up at 128 feet, which I agree with. It's very possible that it would only take 2 seconds to fall that distance, since throwing the rock up has several negative factors: initial inertia, gravity, arm strength, natural arc.

I'll try to figure out the "b" portion of the problem next.

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Jonathan
goodoldtex
goodoldtex
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United States
2005/02/15, 11:26 AM
anni, the easiest way to find the maximum (if you have a graphing calculator ex: TI 83+) is to go into your graph (y=) and put the equation in the first y. hit graph. zoom fit. you'll have to press 2nd then calculate and go to maximum. it will tell you the height (y) and the seconds (x)

i'm assuming the quadratic formula is correct...so i assume the -16x^2 is the force of gravity, because i remember it is -9.8m/s(squared)...but i do not remember the force of gravity in feet. If you have it all correct plug it into the quad form and you'll come out perfectly

sorry i cannot give you a real answer, i leant my graphing calculator to my girlfriend for the semester since i am not taking any math courses.....i hope i helped some.
hecdarec
hecdarec
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United States
2005/02/15, 12:37 PM
Is this a required course? I pray that I never have to do anything like this in my life.

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I skip faster than most people can run.
jonathanweaver
jonathanweaver
Posts: 576
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United States
2005/02/15, 08:34 PM
Ok. And here's the rest of it. Enjoy...

H(x)=-16(x squared) + 64x + 80

a)what is the height of the rock after 3 seconds?
I came up with 128 feet

-16*9 + 64*3 + 80 = -144 + 192 + 80 = 48 + 80 = 128
correct

b) after how many seconds will the rock be at it's highest, and what will that height be?

rock will be at highest when v=0. Velocity = h'(x)
Velocity = (-16(x squared) + 64x + 80)'
Velocity = -32x + 64
V = 0 at x=2
Height = -64 + 128 + 80 = 144

c)How long will it take the rock to reach the water. Hint the height would be 0) You may use the quadratic formula or the factoring technique.

H(x) = 0 = -16(x squared) + 64x + 80
div by 16
0 = -x squared + 4x + 5
x squared = 4x + 5
x = 5 (or -1, but that doesn't fit the word problem)


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Jonathan
Anni313
Anni313
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2005/02/15, 08:38 PM
Jonathan you are THE best! That's what I came up with after I got done agonizing over why I was being made to this.

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Anni

*******
In my head, I am out of my mind....
Anni313
Anni313
Posts: 1,790
Joined: 2004/03/04
United States
2005/02/15, 08:46 PM
Jonathan! Check your email!

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Anni

*******
In my head, I am out of my mind....
jonathanweaver
jonathanweaver
Posts: 576
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United States
2005/02/16, 10:21 AM
No problem, Anni. I LOVE Algebra!!! My wife thinks I should teach lower lever Algebra in college. I would teach up through College Algebra. After that, it starts getting difficult. LOL

When I was in high school, my cousin, a CS major in college at the time, told me, "I have already forgotten more math than you've learned." I was able to tell my sister-in-law that the other night while helping her with her high school Algebra teacher.

For fun, I bought "The Mensa Puzzle Calendar". I get all the math ones right, and most of the word ones wrong. I even found a mistake on one of the math questions. I had a friend double-check it, and he agreed with me.

Bring on the Algebra problems...

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Jonathan
nerraw
nerraw
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United States
2005/02/16, 10:58 AM
And you posted that on Valentines day too!
Anni313
Anni313
Posts: 1,790
Joined: 2004/03/04
United States
2005/02/16, 06:52 PM
I really appreciate your help Jonathan. I have tremendous respect for people who understand algebra because I absolutely do NOT get it.

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Anni

*******
In my head, I am out of my mind....
7707mutt
7707mutt
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United States
2005/02/16, 06:59 PM
You math Nerds LOL that gave me a headache!

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If you can itch your nose after arm day...do another set!

7707mutt@freetrainers.com
jonathanweaver
jonathanweaver
Posts: 576
Joined: 2004/06/14
United States
2005/02/17, 12:15 PM
I'd trade half of my math skills for half of your muscle mass, Mutt.

I tell my wife stuff like that. I'd trade half my metabolism and energy for half of her muscle and fat mass. (I can burn off the fat, she can't.)

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Jonathan
jonathanweaver
jonathanweaver
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Joined: 2004/06/14
United States
2005/02/18, 01:23 PM
Ok. At a beauty parlor one busy Saturday, the staff found that there were 10 blondes. One-fourth of the customers were redheads, and half were brunettes.

How many customers altogether?

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Jonathan
hecdarec
hecdarec
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2005/02/18, 01:25 PM
10

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I was 17 years old the first time my father told me he loved me.
jonathanweaver
jonathanweaver
Posts: 576
Joined: 2004/06/14
United States
2005/02/18, 01:31 PM
Nope. There we 10 blondes. There were also some redheads and brunettes in there as well. Good guess though. :-)

I figured DFly would enjoy this one...

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Jonathan
howdiekat
howdiekat
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2005/02/18, 01:33 PM
40

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if you get me started on quentin tarantino films, chances are you won't be able to shut me up.

excuses are really good for making you fat.

margarine is a liar who announces, "i am butter!"
hecdarec
hecdarec
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United States
2005/02/18, 01:45 PM
40

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I was 17 years old the first time my father told me he loved me.
jonathanweaver
jonathanweaver
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Joined: 2004/06/14
United States
2005/02/18, 03:04 PM
Yup.

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Jonathan
Carivan
Carivan
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Canada
2005/02/18, 03:30 PM
40?

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Scales are for dead weight: We are not dead yet!
Still trying to find out how to do the Hollywood Free Press.

Ivan
carivan@freetrainers.com
Montreal Canada
hecdarec
hecdarec
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United States
2005/02/18, 04:53 PM
Carivan if you cannot understand the answer I am not going to waste my time explaining it to you.

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I was 17 years old the first time my father told me he loved me.
Carivan
Carivan
Posts: 8,542
Joined: 2002/01/20
Canada
2005/02/18, 05:30 PM
C'mon Hec, explain it to me!

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Scales are for dead weight: We are not dead yet!
Still trying to find out how to do the Hollywood Free Press.

Ivan
carivan@freetrainers.com
Montreal Canada