T TheRelevant Full Member 7+ Year Member Joined Jun 14, 2014 Messages 62 Reaction score 37 Points 4,651 Pre-Medical Jun 24, 2014 #1 Advertisement - Members don't see this ad My algebra's a little rusty, but there's no way this can be right, is there? Subbing random numbers in did not seem to work. (2h/2g)^1/2 is not (1/2)^2*(2h/g)
Advertisement - Members don't see this ad My algebra's a little rusty, but there's no way this can be right, is there? Subbing random numbers in did not seem to work. (2h/2g)^1/2 is not (1/2)^2*(2h/g)
el_duderino Some men play tennis, I erode the human soul Verified Member 10+ Year Member Physician Joined Jan 26, 2012 Messages 6,232 Reaction score 7,580 Points 6,816 Resident [Any Field] Jun 24, 2014 #2 I think the typo is that the square root symbol should extend over the (2h/g) and (t) as well. Then it works out. If t = 2h/g, then: (2h/2g)^(1/2) = (t/2)^(1/2) Upvote 0 Downvote
I think the typo is that the square root symbol should extend over the (2h/g) and (t) as well. Then it works out. If t = 2h/g, then: (2h/2g)^(1/2) = (t/2)^(1/2)
T TheRelevant Full Member 7+ Year Member Joined Jun 14, 2014 Messages 62 Reaction score 37 Points 4,651 Pre-Medical Jun 24, 2014 #3 Awesome! That's what I thought. Was starting to panic at my lack of algebraic competency 😛 Upvote 0 Downvote