Posted: 1/13/2002 4:36:35 AM EST
I am doing a math project.
a quadratic equation is written as:
y = ax2(should be superscript)+bx+c
anyway I thought if it involved gravity the value for A would become 9.8 but my math book says 4.9, why is this?



I believe you are thinking of velocity as a function of time due to gravity. It is actually a definite integral of "a" (acceleration) over the interval from zero to "t" (time), and works out to be:
1/2 a tsquared
Since "a" is 9.8 m/secsquared, "1/2 a" would be 4.9 m/secsquared.
This is a guess, though, since you are quoting the math book out of context.



Relax, CGN. 2 plus 2 equals 5, and all letter variables have the value that you need to complete the equation with ease.
Did you buy your 1984 calendar yet?
Double(OrNothing


That's PERSEC, not OPSEC. AvengeR15

Originally Posted By CanadianGunNut:
I am doing a math project.
a quadratic equation is written as:
y = ax2(should be superscript)+bx+c
anyway I thought if it involved gravity the value for A would become 9.8 but my math book says 4.9, why is this? View Quote 

I am not the means to any end others may wish to accomplish. I am not a tool for their use. I do not surrender my treasures to be flung to the winds as alms for the poor.
~ From "Anthem" 
Um...I thought the quadratic equation was A(squared)+B(squared)=c{squared}?
It is the relationship of the three sides of a triangle where one angle is 90degrees,c being the longest side. Once you have the lenghts of the sides figured out you can then resolves the angles since you already know that one is 90 degrees and the total sum of angles if 180 degrees.
Sgtar15



Originally Posted By sgtar15:
Um...I thought the quadratic equation was A(squared)+B(squared)=c{squared}?
It is the relationship of the three sides of a triangle where one angle is 90degrees,c being the longest side. Once you have the lenghts of the sides figured out you can then resolves the angles since you already know that one is 90 degrees and the total sum of angles if 180 degrees.
Sgtar15 View Quote 

I am not the means to any end others may wish to accomplish. I am not a tool for their use. I do not surrender my treasures to be flung to the winds as alms for the poor.
~ From "Anthem" 
Sounds like position as a function of time in the vertical plane for an object being acted upon by a constant acceleration (ie gravity).
9.8M/s^2 is the force of gravity on earth, give or take. '4.9' (UNITS!!  calculating and including the full units in your equations allows you to double check for any errors) could be '1/2a' as marvl points out. Or it could be 'a' on another planet, a favorite topic for some math professors who seldom get invited places... :)
The full equation and its derivation is at
[URL]http://stravinsky.ucsc.edu/~josh/5A/book/notes/node22.html[/URL]
Edited to make link active



Originally Posted By marvl:
I believe you are thinking of velocity as a function of time due to gravity. It is actually a definite integral of "a" (acceleration) over the interval from zero to "t" (time), and works out to be:
1/2 a tsquared
Since "a" is 9.8 m/secsquared, "1/2 a" would be 4.9 m/secsquared.
This is a guess, though, since you are quoting the math book out of context. View Quote 


Ouch:
Now my head hurts !!!
Jay [>:/]



Here's my weak, halfasleep effort.
Since you mention gravity and quadratic equations, it sounds like you are studying parabolic trajectories.
The quadratic equation will describe the trajectory, while the downward acceleration of gravity will be controlled by the 9.81m/s^2.
So,
r = r(initial) + v*t  1/2 g t^2
The 2 denominator in the above equation accounts for your question of why the acceleration of gravity is half the value you think it ought to be.
"Motion with Constant Acceleration", from Resnick/Halliday/Krane



Damn math teachers practicing physics without a license. Why do they insist on confusing things by not using units, using the letter x to stand for time instead of t, and using a to stand for 1/2 acceleration...
Anyway y=ax^2+bx+c
y=position as function of time
x=time in seconds
a=1/2 acceleraton=4.9meters/sec^2
b=velocity at time 0
c=position at time 0
For example, you're standing on the roof at a height of 4 meters on New Years Eve. You shoot your AR straight up in the air and it has a muzzle velocity of 900meters/sec. How high will the bullet be after one second?
a=4.9, b=900, c=4, x=1, solve for y
y=4.9*1+ 900*1 + 4 = 899.1 meters.



Originally Posted By marvl:
I believe you are thinking of velocity as a function of time due to gravity. It is actually a definite integral of "a" (acceleration) over the interval from zero to "t" (time), and works out to be:
1/2 a tsquared
Since "a" is 9.8 m/secsquared, "1/2 a" would be 4.9 m/secsquared.
This is a guess, though, since you are quoting the math book out of context. View Quote 


Next time Lockheed Martin does some engineering work on a space shuttle, I'm gonna recommend you fellas! [:E]



Great idea ECS,
that way we can finally get that poor old VentureStar back on track. It's gonna take more than familiarity with the quadratic formula to solve its problems :)



AR15.COM is the worldâ€™s largest firearm community and is a gathering place for firearm enthusiasts of all types.
From hunters and military members, to competition shooters and general firearm enthusiasts, we welcome anyone who values and respects the way of the firearm.
Subscribe to our monthly Newsletter to receive firearm news, product discounts from your favorite Industry Partners, and more.
Copyright © 19962017 AR15.COM LLC. All Rights Reserved.
Any use of this content without express written consent is prohibited.