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3.14159265...
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Postby 3.14159265... » Thu Jun 21, 2007 3:08 am UTC

I give you a cannon that shoots a ball of mass m at velocity v.

At a distance d from the tip of the canon is an infinite vertical wall.

You shoot the ball, and it hits the wall h meters above the point horizontal from the tip of the canon.

You are on a planet with gravitational acceleration of a on the surface.

P=h-t, where P is the points you scored.

What angle should you put your canon at to score maximum points.
Note:
1) points will almost in all cases be negative.
2) assume d>1m for simplicity.
3) assume the affects of height are negligable on a

4) For a harder problem let P=h-t+v', where v' is the final velocity.
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Postby SpitValve » Thu Jun 21, 2007 3:36 am UTC

Are we allowed to vary v? or just the angle?

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Postby ArmonSore » Thu Jun 21, 2007 3:55 am UTC

Let's see if this works.

h = -.5*a*t^2 + v*sin(theta)*t
d = v*cos(theta)*t
t = d/(v*cos(theta))

P = h - t
P = -.5*a*t^2 + v*sin(theta)*t - t
P = -.5*a*(d^2) /(v^2)*(cos(theta)^2) + v*sin(theta)*(d/v*cos(theta) -(d/v*cos(theta))

Taking the derivative of P:
dP/dtheta = -*a*(d/v)^2 * sin(theta)/cos(theta)^3 + d*sec(theta)^2 - (d/v)*sin(theta)*(sec(theta)^2)

Setting the derivative equal to zero, and simplifying, in order to find the max or min:
0 = -(a*d/v)* tan(theta) + 1 - sin(theta)

1 = (a*d/v)*tan(theta) + sin(theta)

Now comes the guess work:
0 <= theta <= 90
sin and tan are positive on that interval. It is reasonable to assume that theta is small since both tan and sin are positive and are adding together. This assumption is more reasonable if the quantity (a*d/v) happens to be large.

For small angles tangent is approximately sine:
1 = (a*d/v)*sin(theta) + sin(theta))
1 = sin(theta)*((a*d/v) + 1)
sin(theta) = 1/((a*d/v)+1)

theta = arcsin(1/((a*d/v)+1))

That would be my attempt.


Note: I still haven't proven that I have a relative maximum here, and not a relative min.

P'' = -(ad/v)*sec(theta)^2 - cos(theta)
Both cosine and secant-squared are positive on the interval from 0 to 90. Therefore the second derivative is always negative on this interval, and my above guess is an approximation to the position of a relative max.
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Postby 3.14159265... » Thu Jun 21, 2007 4:41 am UTC

SpitValve wrote:Are we allowed to vary v? or just the angle?

only the angle.

ArmonSore: Thats pretty much where I got to, and couldn't not find a way to find theta without the use of approximating off a graph assuming theta is not small.
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Postby ArmonSore » Thu Jun 21, 2007 5:10 am UTC

3.14159265... wrote:ArmonSore: Thats pretty much where I got to, and couldn't not find a way to find theta without the use of approximating off a graph assuming theta is not small.


Since explicit numbers weren't given I was kind of assuming that the distances, velocities, and acceleration would be random for each firing. Plugging in numbers and using a graphing calculator would be a lot simpler.

I just realized that the answer is bothering the heck out of me.

1 = (a*d/v)*tan(theta) + sin(theta)

Since tangent and sine approach zero as theta approaches zero we can always find a theta that works no matter how large the quantity (a*d/v) is. But we could imagine a d sufficiently large or a v sufficiently small, such that there should be no theta where the ball even reaches the wall.

I imagine this is because the equations I set up implicitly assumed that d and v were in the proper proportions. It would be nice if my above answer reflected the possibility that there is no solution.

Anybody have an idea on how to solve the problem so that it will tell us if there is a solution, and then give us an explicit answer?
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Postby 3.14159265... » Thu Jun 21, 2007 5:18 am UTC

I don't think it will happen for any thetha under 90, where neither a, d, and v are infinitely large or infinitely small.

See thats what is bothering me too, I can't find an explicit solution. Another problem I have been tackling is.

let c, a, and b be constants:

c = (a/sin(x))+(b/cos(x)), solve for x. For simplicity assume a and/or b to equal 1
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Postby Cosmologicon » Thu Jun 21, 2007 5:40 am UTC

I made the substitution x = tan(theta) and got the condition down to a fourth-degree polynomial on x:

(1 + x^2) (v^2 - adx)^2 - v^2 x^2 = 0

So there you have it. A fourth-degree polynomial that doesn't factor symbolically. I don't know why you'd expect to be able to find an explicit solution. There are lots of equations you just can't solve explicitly, especially if you throw in lots of constants.

The one obvious simplification is for v >> 1, you can ignore the v^2 x^2 term, and then the other factors give you x = v^2 / ad. This is effectively just maximizing h and ignoring the -t part of the points. In this regime, the times involved are small compared with the distances. It's sort of an artifact of the fact that you're subtracting a time from a distance without any conversion factor.

For example, for v = 100, d = 100, a = 10, I get that there's a maximum of P = 485.000 at x = 9.9005. Anyway, clearly for many parameters you can't assume that x is small.

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Postby bitwiseshiftleft » Thu Jun 21, 2007 2:56 pm UTC

Cosmologicon wrote:So there you have it. A fourth-degree polynomial that doesn't factor symbolically. I don't know why you'd expect to be able to find an explicit solution. There are lots of equations you just can't solve explicitly, especially if you throw in lots of constants.


Quartic polynomials are always exactly soluble by roots. In fact, there's a quartic formula. It's nasty, though.

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Postby Cosmologicon » Thu Jun 21, 2007 3:05 pm UTC

bitwiseshiftleft wrote:
Cosmologicon wrote:So there you have it. A fourth-degree polynomial that doesn't factor symbolically. I don't know why you'd expect to be able to find an explicit solution. There are lots of equations you just can't solve explicitly, especially if you throw in lots of constants.


Quartic polynomials are always exactly soluble by roots. In fact, there's a quartic formula. It's nasty, though.

If you count having cosines in your solution, then yeah it's always soluble. In general you can't solve it without taking the cosine of something numerically, which for practical purposes isn't any better than just finding the root numerically. That's why I'm not satisfied with the cubic or quartic formula, but I should have been clearer on what I meant by "no explicit solution".

EDIT: Actually, that's not fair.... If I'd gotten a solution that involved a cosine but not the quartic formula, I probably would have been happy with it. I can't really say why I don't consider using the quartic formula to be an explicit solution, but I don't.

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Postby gmalivuk » Thu Jun 21, 2007 5:03 pm UTC

Cosmologicon wrote:
bitwiseshiftleft wrote:Quartic polynomials are always exactly soluble by roots. In fact, there's a quartic formula. It's nasty, though.

If you count having cosines in your solution, then yeah it's always soluble.


Are you sure you need cosines?

As far as I can tell, one of the solutions is "simply"
v^2/(2*a*d) + (Sqrt[(-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/ (a^2*d^2) + 2*((-5*((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2)))/6 + ((3*v^8)/(16*a^4*d^4) - (v^4*(a^2*d^2 - v^2 + v^4))/(4*a^4*d^4) - ((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/ (a^2*d^2))^2/12)/ (3*((-((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))^3/108 - ((-2*v^2)/(a*d) - v^6/(a^3*d^3) + (v^2*(a^2*d^2 - v^2 + v^4))/(a^3*d^3))^ 2/8 + (((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))*( (-3*v^8)/(16*a^4*d^4) + (v^4*(a^2*d^2 - v^2 + v^4))/(4*a^4* d^4)))/3)/2 + Sqrt[ ((3*v^8)/(16*a^4*d^4) - (v^4*(a^2*d^2 - v^2 + v^4))/(4*a^4*d^4) - ((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))^2/12)^3/27 + (-((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))^3/108 - ((-2*v^2)/(a*d) - v^6/(a^3*d^3) + (v^2*(a^2*d^2 - v^2 + v^4))/(a^3* d^3))^2/8 + (((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))* ((-3*v^8)/(16*a^4*d^4) + (v^4*(a^2*d^2 - v^2 + v^4))/(4*a^4* d^4)))/3)^2/4])^(1/3)) - ((-((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))^3/108 - ((-2*v^2)/(a*d) - v^6/(a^3*d^3) + (v^2*(a^2*d^2 - v^2 + v^4))/(a^3*d^3))^2/ 8 + (((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))* ((-3*v^8)/(16*a^4*d^4) + (v^4*(a^2*d^2 - v^2 + v^4))/(4*a^4*d^4)))/3)/2 + Sqrt[((3*v^8)/(16*a^4*d^4) - (v^4*(a^2*d^2 - v^2 + v^4))/(4*a^4*d^4) - ((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))^2/12)^3/27 + (-((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))^3/108 - ((-2*v^2)/(a*d) - v^6/(a^3*d^3) + (v^2*(a^2*d^2 - v^2 + v^4))/(a^3* d^3))^2/8 + (((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))* ((-3*v^8)/(16*a^4*d^4) + (v^4*(a^2*d^2 - v^2 + v^4))/(4*a^4* d^4)))/3)^2/4])^(1/3))] + Sqrt[-3*((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2)) - 2*((-5*((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2)))/6 + ((3*v^8)/(16*a^4*d^4) - (v^4*(a^2*d^2 - v^2 + v^4))/(4*a^4*d^4) - ((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/ (a^2*d^2))^2/12)/ (3*((-((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))^3/108 - ((-2*v^2)/(a*d) - v^6/(a^3*d^3) + (v^2*(a^2*d^2 - v^2 + v^4))/(a^3*d^3))^ 2/8 + (((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))*( (-3*v^8)/(16*a^4*d^4) + (v^4*(a^2*d^2 - v^2 + v^4))/(4*a^4* d^4)))/3)/2 + Sqrt[ ((3*v^8)/(16*a^4*d^4) - (v^4*(a^2*d^2 - v^2 + v^4))/(4*a^4*d^4) - ((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))^2/12)^3/27 + (-((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))^3/108 - ((-2*v^2)/(a*d) - v^6/(a^3*d^3) + (v^2*(a^2*d^2 - v^2 + v^4))/(a^3* d^3))^2/8 + (((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))* ((-3*v^8)/(16*a^4*d^4) + (v^4*(a^2*d^2 - v^2 + v^4))/(4*a^4* d^4)))/3)^2/4])^(1/3)) - ((-((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))^3/108 - ((-2*v^2)/(a*d) - v^6/(a^3*d^3) + (v^2*(a^2*d^2 - v^2 + v^4))/(a^3*d^3))^2/ 8 + (((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))* ((-3*v^8)/(16*a^4*d^4) + (v^4*(a^2*d^2 - v^2 + v^4))/(4*a^4*d^4)))/3)/2 + Sqrt[((3*v^8)/(16*a^4*d^4) - (v^4*(a^2*d^2 - v^2 + v^4))/(4*a^4*d^4) - ((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))^2/12)^3/27 + (-((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))^3/108 - ((-2*v^2)/(a*d) - v^6/(a^3*d^3) + (v^2*(a^2*d^2 - v^2 + v^4))/(a^3* d^3))^2/8 + (((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))* ((-3*v^8)/(16*a^4*d^4) + (v^4*(a^2*d^2 - v^2 + v^4))/(4*a^4* d^4)))/3)^2/4])^(1/3)) - (2*((-2*v^2)/(a*d) - v^6/(a^3*d^3) + (v^2*(a^2*d^2 - v^2 + v^4))/(a^3*d^3)))/ Sqrt[(-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2) + 2*((-5*((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2)))/6 + ((3*v^8)/(16*a^4*d^4) - (v^4*(a^2*d^2 - v^2 + v^4))/(4*a^4*d^4) - ((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))^2/12)/ (3*((-((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))^3/108 - ((-2*v^2)/(a*d) - v^6/(a^3*d^3) + (v^2*(a^2*d^2 - v^2 + v^4))/(a^3* d^3))^2/8 + (((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))* ((-3*v^8)/(16*a^4*d^4) + (v^4*(a^2*d^2 - v^2 + v^4))/(4*a^4* d^4)))/3)/2 + Sqrt[ ((3*v^8)/(16*a^4*d^4) - (v^4*(a^2*d^2 - v^2 + v^4))/(4*a^4*d^4) - ((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))^2/12)^3/27 + (-((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))^3/108 - ((-2*v^2)/(a*d) - v^6/(a^3*d^3) + (v^2*(a^2*d^2 - v^2 + v^4))/(a^3* d^3))^2/8 + (((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))* ((-3*v^8)/(16*a^4*d^4) + (v^4* (a^2*d^2 - v^2 + v^4))/(4*a^4*d^4)))/ 3)^2/4])^(1/3)) - ((-((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))^3/108 - ((-2*v^2)/(a*d) - v^6/(a^3*d^3) + (v^2*(a^2*d^2 - v^2 + v^4))/(a^3* d^3))^2/8 + (((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))* ((-3*v^8)/(16*a^4*d^4) + (v^4*(a^2*d^2 - v^2 + v^4))/(4*a^4* d^4)))/3)/2 + Sqrt[ ((3*v^8)/(16*a^4*d^4) - (v^4*(a^2*d^2 - v^2 + v^4))/(4*a^4*d^4) - ((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))^2/12)^3/27 + (-((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))^3/108 - ((-2*v^2)/(a*d) - v^6/(a^3*d^3) + (v^2*(a^2*d^2 - v^2 + v^4))/(a^3* d^3))^2/8 + (((-3*v^4)/(2*a^2*d^2) + (a^2*d^2 - v^2 + v^4)/(a^2*d^2))* ((-3*v^8)/(16*a^4*d^4) + (v^4*(a^2*d^2 - v^2 + v^4))/(4*a^4* d^4)))/3)^2/4])^(1/3))]])/2

See? No trig at all. Just nice radicals and exponents.

Or, for a=9.8, d=10, and v=100, about 103.061
101.02 is the other real root.
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Postby Cosmologicon » Thu Jun 21, 2007 5:47 pm UTC

Yes, I'm sure you need cosines. It's called Casus Irreducibilis. Even if your solution is real, you have to take the square root of a complex number a couple of times, and the imaginary parts of it just cancel out. The only way to express it with real math is by converting the complex numbers to polar, ie, use trig.

I first found out about it when I wanted to express the cosine of 10 degrees using just radicals. I found a cubic that cos(10 deg) satisfies using the triple-angle formula. Imagine my chagrin when I found that applying the cubic formula and simplifying leads to an answer of cos(10 deg).

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Postby gmalivuk » Thu Jun 21, 2007 6:50 pm UTC

Cosmologicon wrote:Yes, I'm sure you need cosines. It's called Casus Irreducibilis. Even if your solution is real, you have to take the square root of a complex number a couple of times, and the imaginary parts of it just cancel out. The only way to express it with real math is by converting the complex numbers to polar, ie, use trig.

I first found out about it when I wanted to express the cosine of 10 degrees using just radicals. I found a cubic that cos(10 deg) satisfies using the triple-angle formula. Imagine my chagrin when I found that applying the cubic formula and simplifying leads to an answer of cos(10 deg).


I imagine the problem must actually arise when you take the cube root of a complex number, since
Image
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Postby Cosmologicon » Thu Jun 21, 2007 7:05 pm UTC

Yeah, that sounds right now that you mention it. It is the cube roots, at least in the cubic equation.

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Postby gmalivuk » Thu Jun 21, 2007 7:20 pm UTC

Cosmologicon wrote:Yeah, that sounds right now that you mention it. It is the cube roots, at least in the cubic equation.


Of course, you do have to take some cube roots in the quartic formula, of numbers which may not be real. So yeah, you're right, you'd need some trig in there. In particular, with the values I gave above, you get

Code: Select all

2500/49 +
 (Sqrt[-12500099/2401 + 2*(62500495/14406 +
       (624995050009801*(-1)^(2/3))/(207532836*
         (15624822479110074029701/2989718035416 -
           (62500*Sqrt[15624818427422574029701/3])/
            847425747)^(1/3)) -
       (-15624822479110074029701/2989718035416 +
         (62500*Sqrt[15624818427422574029701/3])/
          847425747)^(1/3))] +
   Sqrt[37500297/2401 - 2*(62500495/14406 +
       (624995050009801*(-1)^(2/3))/(207532836*
         (15624822479110074029701/2989718035416 -
           (62500*Sqrt[15624818427422574029701/3])/
            847425747)^(1/3)) -
       (-15624822479110074029701/2989718035416 +
         (62500*Sqrt[15624818427422574029701/3])/
          847425747)^(1/3)) + 49010000/
      (117649*Sqrt[-12500099/2401 +
         2*(62500495/14406 + (624995050009801*
             (-1)^(2/3))/(207532836*
             (15624822479110074029701/2989718035416 -
               (62500*Sqrt[15624818427422574029701/3])/
                847425747)^(1/3)) -
           (-15624822479110074029701/2989718035416 +
             (62500*Sqrt[15624818427422574029701/3])/
              847425747)^(1/3))])])/2
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Postby dumbclown » Thu Jun 21, 2007 7:51 pm UTC

I don't think there is enough information to work out a maximum. Does the ball have to hit the wall for it to count as a score? If not then theta = 0 will probably be the maximum. If it does need to hit then you will need to assumption that v is large enough so that at the maximum trajectory of the ball is larger then d. This will mean there will be two values of theta for which the ball will hit the exact base of the wall (the maximum should lie between these two angles).

Using ArmonSore starting equations

theta = 0.5 arcsin(a*d/v^2)

Going on that equation you need more information on the relation of v to d.

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Postby ArmonSore » Thu Jun 21, 2007 9:11 pm UTC

dumbclown wrote: If it does need to hit then you will need to assumption that v is large enough so that at the maximum trajectory of the ball is larger then d. This will mean there will be two values of theta for which the ball will hit the exact base of the wall (the maximum should lie between these two angles).


It does need to hit the wall. And the above equations implicitly assume that v is large enough such that the ball will transverse a distance d. The fact that the same time parameter is used in each equation implies that when the the ball reaches the distance d it must be at height h. This is a statement the ball must reach the wall.

To find out what those conditions must be, let's use the range equation(stated without proof because I'm lazy):

R = ((v^2/a)) *sin(2theta)

This equation is maximized when sin(2*theta) = 1
Maximum horizontal distance:
R = (v^2)/a
Assumedly:
d <= R
d <= (v^2)/a
(a*d)/(v^2) >= 1

By the way, I just discovered a mistake in my above work:
0 = -*a*(d/v)^2 * sin(theta)/cos(theta)^3 + d*sec(theta)^2 - (d/v)*sin(theta)*(sec(theta)^2)

simplifies to:

1 = (a*d/(v^2))*tan(theta) + (1/v)*sin(theta)

NOT: 1 = (a*d/v)*tan(theta) + sin(theta)

So the equation works as long as the term in front of tangent happens to be greater than one. That's nice to know.
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Postby dumbclown » Fri Jun 22, 2007 8:26 am UTC

Should be a negative in front of the (a*d/(v^2)) to. The sign should have changed for the derivative.

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Postby 3.14159265... » Fri Jun 22, 2007 4:20 pm UTC

I let mine be positive because a is -9.81, so it makes it easier to cancel the signs.
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