We can solve the problem by using the conservation of energy.
In fact, in order to reach an altitude of h above the ground, the initial kinetic energy of the material must be equal to the final gravitational potential energy at altitude h:
![(1)/(2)mv^2 = mgh](https://img.qammunity.org/2019/formulas/physics/college/5wgg5fyvdsa0s4ajrbo9698npp9kpbsc0w.png)
where the term on the left is the kinetic energy, and the term on the left is the gravitational potential energy, and where
m is the mass of the material
v is its initial velocity
g is the gravitational acceleration on Io
h is the altitude
If we use the data of the problem:
![h=500 km = 5 \cdot 10^5 m](https://img.qammunity.org/2019/formulas/physics/college/jdhj429qcrj39vflw60gujhyz9drn2lxrs.png)
![g=1.81 m/s^2](https://img.qammunity.org/2019/formulas/physics/college/v6nm34ilo9cjq3264pg9yg35asr9sn0w93.png)
And we re-arrange the formula, we find the velocity the material should have in order to reach the altitude of 500 km:
![v= √(2gh) = √(2(1.81 m/s^2)(5 \cdot 10^5 m)) =1345 m/s](https://img.qammunity.org/2019/formulas/physics/college/ibls80t3ylrs74e2813bl2nxh8xzijn5na.png)