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:
[tex] \frac{1}{2}mv^2 = mgh [/tex]
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:
[tex]h=500 km = 5 \cdot 10^5 m[/tex]
[tex]g=1.81 m/s^2[/tex]
And we re-arrange the formula, we find the velocity the material should have in order to reach the altitude of 500 km:
[tex]v= \sqrt{2gh} = \sqrt{2(1.81 m/s^2)(5 \cdot 10^5 m)} =1345 m/s[/tex]
