So, a number of my team worked on the shooter today at our Lead Technical Mentor’s house. I couldn’t be there. When I asked the length of the shots that were made, I was given this and so now I give it to Chief Delphi. The length of the shot is W.
Ha, calculus, Good times, good times, oh wait i’m still learning that stuff. I’d try to solve it but I’m to lazy right now.
Someone’s messing with you on that equation.
Wow, that’s pretty far (assuming the answer is in feet?). Do you happen to know what size the wheel was and what motor they used?
We are using eight inch pneumatics with 2 Cims. And yes, I know the Cim argument - we’re reserving judgement on changing dependent on how testing goes when we have the full bot assembled and test it.
For the sake of the lazier and less calculus inclined people (I personify both of these categories 
) could someone please post what omega is?
Yeah…I’m in Algebra 2/Trig. This makes no sense to me right now. 
Nice aproximation!
Sig-Figs?
It was 70ft.
In that case, the constant on the left-hand-side should have been 172.551465, not 169.28894.
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I don’t do sig figs.
http://www.av8n.com/physics/uncertainty.htm#sec-more-about-sigfig
I gave an answer to 6 decimal places because that is what is required to match all 5 decimal digits in the problem as stated. See attachment.
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