Step 1: Identify the initial and final positions of the object.
The object is projected from Earth's surface (at distance $R_E$ from Earth's center) with initial speed $V$. It reaches a maximum height of $\frac{4}{5}R_E$ above the surface, so the final distance from Earth's center is: \[ r_f = R_E + \frac{4}{5}R_E = \frac{9}{5}R_E \] At this maximum height, the object momentarily comes to rest (speed = 0).
Step 2: Apply conservation of mechanical energy.
Total mechanical energy is conserved (no friction in space). Initial state: speed $V$ at $r = R_E$. Final state: speed 0 at $r = 9R_E/5$. Thus: \[ \frac{1}{2}mV^2 - \frac{GMm}{R_E} = 0 - \frac{GMm}{\frac{9R_E}{5}} \] \[ \frac{1}{2}V^2 = \frac{GM}{R_E} - \frac{5GM}{9R_E} = \frac{9GM - 5GM}{9R_E} = \frac{4GM}{9R_E} \] \[ V^2 = \frac{8GM}{9R_E} \]
Step 3: Recall the formula for escape velocity.
The escape velocity $V_E$ is the minimum speed needed to escape Earth's gravity from the surface: \[ V_E = \sqrt{\frac{2GM}{R_E}} \implies V_E^2 = \frac{2GM}{R_E} \] This is derived by setting total energy to zero: $\frac{1}{2}V_E^2 - GM/R_E = 0$.
Step 4: Compute the ratio $(V/V_E)^2$.
\[ \left(\frac{V}{V_E}\right)^2 = \frac{V^2}{V_E^2} = \frac{\frac{8GM}{9R_E}}{\frac{2GM}{R_E}} = \frac{8}{9} \times \frac{1}{2} = \frac{8}{18} = \frac{4}{9} \]
Step 5: Take the square root to find $V/V_E$.
\[ \frac{V}{V_E} = \sqrt{\frac{4}{9}} = \frac{2}{3} \] Since speed is positive, we take the positive root.
Step 6: Interpret the result physically.
The object needs only $\frac{2}{3}$ of escape velocity to reach a height of $\frac{4}{5}R_E$. The fact that $V < V_E$ confirms the object cannot escape Earth's gravity and will return after reaching the maximum height. As $V \to V_E$, the maximum height approaches infinity. \[ \boxed{\frac{V}{V_E} = \frac{2}{3}} \]