Recent content by CarlosK

by: Euclides (http://pir2.forumeiros.com/)

I got this v = √ (2gh/(β+1)). I need this v=∆x√(g/(2H))

Sorry.. v = √ (2gh/(β+1)) is correct. Thanks

The exercise asked to deduce these formulas.

Hi All, Question: How to derive these equations of missile launch? v=∆x√(g/2H) v=√(2gh/β+1) I need to use these formulas: ∆y=voyt - (gt²)/2 ∆x=vxt K=(mv²)/2 + (Iω²)/2 ∆UG=mg∆y I=βmr² β=2/5 Thanks, Carlos.