Quote:
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Originally Posted by Mike Rush
Use the energy equations
mgh = 1/2 (mv^2)
gh=(v^2)/2
h=v^2/2g
h=100/64.4
h=1.55 ft
if the object goes up over 1.55 feet it is travelingmore than 10 ft/s
(I hope this is right... It's been a long time  ) |
Er... I am currently taking AP Physics so I am not very experienced with it. But it seems to me that you are saying: Potential Energy = Kinetic Energy... Well, when the robot is hanging it's got PE, but no KE... I think it was lucky that when solving for 'h', you came out with v^2/2g
I used Andy's method of applying the equation "V^2=Vo^2+2aX" and derived the same formula.. Hmmmmm.