I'm now getting 950 cm/s^2, slightly off from your calculation, but much much more realistic. I wonder where the discrepancy is now, but thank you so much for helping me understand.
Hmm, so is (Δt)_avg the average time required for the glider to pass through the photogate from start to finish? That makes more sense, because the photogate couldn't know when the glider was released.
Hi, so this is a lab in which we used an air track at an angle and a glider to gather some data through various trials, ultimately to calculate "g". L_glider = 10.15 cm
x (photogate activation point) = 547.5 mm or 54.75 cm
x_0 (release point) = 1800.0 mm or 180.00 cm
(Δx)_midpoint = | x - x_0 |...