The problem can be solved by applying the principle of conservation of mechanical energy. The initial kinetic energy of the rock is converted into gravitational potential energy at its maximum height.

Let $v_0$ be the initial ejection speed and $h_{max}$ be the maximum height reached. We set the gravitational potential energy to be zero at the ejection point.

Initial state (at ejection):

• Kinetic energy: $K_0 = \frac{1}{2}mv_0^2$
• Potential energy: $U_0 = 0$

Final state (at maximum height):

• Kinetic energy: $K_f = 0$ (the rock is momentarily at rest)
• Potential energy: $U_f = mg_{Io}h_{max}$

By conservation of mechanical energy, $K_0 + U_0 = K_f + U_f$:

$$\frac{1}{2}mv_0^2 + 0 = 0 + mg_{Io}h_{max}$$

The mass $m$ cancels out:

$$\frac{1}{2}v_0^2 = g_{Io}h_{max}$$

Solving for the initial speed $v_0$:

$$v_0 = \sqrt{2g_{Io}h_{max}}$$

Now, we substitute the given values, ensuring consistent units.

• $g_{Io} = 1.80$ m/s²
• $h_{max} = 200$ km $= 200 \times 10^3$ m

$$v_0 = \sqrt{2(1.80 \text{ m/s}^2)(200 \times 10^3 \text{ m})}$$ $$v_0 = \sqrt{720000 \text{ m}^2/\text{s}^2}$$ $$v_0 \approx 848.5 \text{ m/s}$$

Rounding to three significant figures, we get $v_0 = 849$ m/s.

Alternatively, using kinematics: The appropriate kinematic equation for motion under constant acceleration is $v_f^2 = v_0^2 + 2a\Delta y$. Here, $v_f = 0$, $a = -g_{Io}$, and $\Delta y = h_{max}$.

$$0 = v_0^2 - 2g_{Io}h_{max}$$

This yields the same expression: $v_0 = \sqrt{2g_{Io}h_{max}}$.