Answer
The percent difference in the height measurement is 32%.
Work Step by Step
$height = \frac{1}{2}at^2 = \frac{1}{2}(9.8)(t^2)$
$h_1 =\frac{1}{2}(9.8)(2.0)^2 = 19.6 ~m$
$h_2 =\frac{1}{2}(9.8)(2.3)^2 = 25.9 ~m$
The percent difference is $\frac{25.9-19.6}{19.6}\times100\% = 32\%$
