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What the heck is equation 2 supposed to be? I don't blame you for it, but I have to wonder about whoever comes up with this stuff. So I'm going to ignore equation 2 (you did say "any help"), and use the formula for the propagation of absolute error (which involves derivatives, so I hope you're up for that). Somewhere along the way, we may see equation 2 appear, and then at least we'll know what it is.

If x is a function of a, b, c, etc, with absolute errors Δa, Δb, Δc, etc., which are independent, then the absolute error in x is

Δx = √[ (dx/da ∙ Δa)² + (dx/db ∙ Δb)² + (dx/dc ∙ Δ c)² + ... ]

where the derivatives are actually partial derivatives. In your case, g is a function of y and t, so

Δg = √[ (dg/dy ∙ Δy)² + (dg/dt ∙ Δt)² ]

where dg/dy = 2/t²

and dg/dt = -4y/t³

so

Δg = √[ (2/t² ∙ Δy)² + (4y/t³ ∙ Δt)² ]

where I've dropped the negative sign on -4y/t³ because it's not going to matter once the term is squared.

Personally, at this point, I'd calculate a value for Δg, and divide it by my calculated value for g, to get the fractional error, dodging the formula for fractional error completely. But if you really need the formula, divide both sides by g, remembering that g = 2y/t² when you divide through on right.

Maybe it would be better to look at it term by term.

(2/t²) / g = (2/t²) / (2y/t²) = 1/y

(4y/t³) / g = (4y/t³) / (2y/t²) = 2/t

I think we're there.

Δg/g = √[ (1/y ∙ Δy)² + (2/t ∙ Δt)² ] = √[ (Δy / y)² + (2 Δt / t)² ]

But I still have no idea what equation 2 is.

how did you get dg/dy=2/t^2 and dg/dy=-4y/t^2