Answer
91.7 km
Work Step by Step
To simplify things we assume the gravitational field is both constant and radial.
At the maximum height, the velocity $u = 0,$ the acceleration $a = g_{Io}$.
Clearly $g_{Io} = G\frac{m_{Io}}{r_{Io}^2} = (6.67 \times 10^{-11})\frac{(8.93\times 10^{22}}{(1.821\times 10^6)^2} = 1.797 m/s^2$
Using SUVAT ($v^2 = u^2 + 2ay$) . . .
$v=\sqrt{2g_{Io}y} = \sqrt{2(1.797)(5 \times 10^5)} = 1.3405 \times 10^3 m/s$
$\implies \Delta y_{earth} = \frac{v^2 - u^2}{2g_{earth}} = \frac{(1.3405 \times 10^3)^2 - (0)^2}{2(9.8)} = 9.17 \times 10^4 \ m = 91.7 km$.
