Answer
The armadillo goes another $~~0.154~m~~$ higher than $0.544~m$
Work Step by Step
We can find the maximum height $y$ when the speed is zero:
$v^2 = v_0^2+2ay$
$y = \frac{v^2 - v_0^2}{2a}$
$y = \frac{(0)^2 - (3.70~m/s)^2}{(2)(-9.8~m/s^2)}$
$y = 0.698~m$
We can find the additional height the armadillo reaches above $y = 0.544~m$:
$\Delta y = 0.698~m-0.544~m = 0.154~m$
The armadillo goes another $~~0.154~m~~$ higher than $0.544~m$
