Physics education research, electronic materials, and the musings of Christopher Moore, Ph.D.
Physics education research, electronic materials, and the musings of Christopher Moore, Ph.D.
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Hi so I'm not sure how much the time for the horizontal range is assumed to be in projectile motion.
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It's simply or
Last edited by Walter Lewin (2010-05-30 10:56:44)
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But perhaps, using trig answers might confuse the OP. He may have required a much more simpler derivation.
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Y=UT+1/2gT^2 (IN VERTICAL DIRECTION ) g=g(effectiv) U=Uy, Y=net vertical displacement from projection point
above eque. in vector form ......this formula is alwags true...T=time period for horigental rang
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T=(2USin*)/g is not alwage true
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