Quote:
Originally posted by Mike F+May 16 2005, 09:22 PM--></div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td>QUOTE (Mike F @ May 16 2005, 09:22 PM)</td></tr><tr><td id='QUOTE'> <!--QuoteBegin-fotze@May 16 2005, 07:48 PM
5=(0)t+0.5at^2
t=2 solve for a
a= 2.5
there is not enough info if the acceleration is dependant on distance form the earth
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That can't be right, because you don't account for the distance travelled during the first second.
OK here you go:
a = (Vf - Vi)/t
Vi=0
Vf -- you know it moves 5m during the second second, so 5m/s is it's
average velocity during the second second, which means it's actual velocity is 5m/s
half way through the second second. Therefore:
a = (5m/s - 0)/1.5s
a = 5m/s/1.5s
a = 3.33m/s/s [/b][/quote]
fata!!!!
Dude, good work. I was looking at it last night and must have spent about 30 minutes on it and didnt get it.
Like fotze, I have been out of school for too long and am a idiot.
And that has only been 13 months...