posted
Well if you told me how you encrypted it I could decrypt it...
Seriously, I'm no good at decoding anything. And that includes subtle hints from females, things like shooting me is probably going to go right over my head. If not I probably wont remember anyways.
posted
Speaking of weird transcendental constants...I have a question for any of the geeky here particularly skilled in the areas of math and/or physics...any takers?
Posts: 88 | Registered: Jan 2003
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posted
Note: this is from a Physics Exam I just took, i have turned it in, so im not cheating or anything like that.
OK, You shoot a ball of mass m, verticaly with velocity 2*Vt, where Vt is the terminal velocty. air resistance is linear such that, Force = -bv and , where b is some constant of drag.
In terms of m, b and g(gravity),
a)find, the Time it takes the ball to travel to its apex,
b)find the height H of the apex
c)find the velocity with which it hits the ground after falling down
What I found interesting is that then solving part c, the ratio between the time up, and the time down, is constant, independent of m, b, or g and the constant is a non terminating, non repeating decimal,(or so it would seem) I was wondering, if anyone else could find the exact value of this constant? like, the natural log of something divided by the natural log of something else?
Stuff to help you if you try, V(t) = Vt*(1-e^(-bt/m))
^ | | Velocity as a function of time, for linear air resistance.
posted
oh yea, the constant turns out to be around, 1.568184792... If any of you know that is off the top of your head, so you dont have to work it all out (And, no its not Pi/2, its close, but not really)
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posted
meltingsnowman, Your solution doesn't work in the limit where b = 0 (no drag). In this case the ration of the time up to the time down = 1. As the drag coefficient increases, this has to change.
Posts: 12591 | Registered: Jan 2000
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posted
Ah, I see part of my problem. The initial velocity of the ball is twice the terman velocity. As b goes to zero, the terminal velocity will approach infinity. As a result the time up and the time down will also approach infinity making the ratio of the two an indeterminate form.
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posted
meltingsnowman, Your answer doesn't work because the sign of the gravitational force doesn't change between when the ball is going up and when the ball is coming down, but the sign of the drag force does change when the ball changes directions. When the ball is going up the solution should be
V(t) = Vt(1-3e^(-bt/m))
When the ball is coming down the solution should be
V(t) = Vt(1-e^(-b(t-tup)/m))
where tup is the time it takes the ball to reach the appex or m ln(3)/b.
so that gives me the height at the apex equals
x = (m^2g/b^2)(ln(3)+2/3);
and I can't get an explicit solution for the time down.
[ September 30, 2003, 01:56 PM: Message edited by: The Rabbit ]
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posted
[interjection]Annie, did you know that in Ivan's apartment at Ricks they had the walls of the living room totally covered with white paper? That way, anyone could write whatever they wanted to on the walls. It was chief flirting spot and there was a lifesize cardboard cutout of John Wayne...but the coolest thing was, along the ceiling, like a border, Mike and Ivan were always in the process of writing the digits of pi. It got longer and longer and it was hilarious. Sorry. This thread just reminded me of that. [/interjection]
Posts: 6415 | Registered: Jul 2000
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posted
Annie, how did I miss that? You're hilarious and oh so creative. Somebody a whole lot cooler is going to appreciate that one of these days. :giggles to self:
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Annie, My younger sister once made square pies for a high school math club party with πr^2 written carved in the top crust.
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posted
Im still very confident of my solution, i checked it many many times for many many hours. The time up you got was right, but the distance wasnt, perhaps you integrated wrong? It should have been H = g(m/b)^2 * (2-ln(3))
He left the time down as an equation you sould solve for T when you had definite values to plug in. I found (by running serveral differnt starting varibles) that the time up and time down were proportional by that constant
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