Answer:
42.96 km/s
Step-by-step explanation:
From the conservation of Energy
![(PE+KE)_i=(PE+KE)_f\\\Rightarrow -(GmM)/(R)+(1)/(2)mv_i^2=0+(1)/(2)mv_f^2](https://img.qammunity.org/2020/formulas/physics/college/tz4er4psm165hzpd2kb5mnqn5timsq0glw.png)
Mass gets cancelled
![-(GM)/(R)+(1)/(2)v_i^2=0+(1)/(2)v_f^2\\\Rightarrow -2(GM)/(R)+v_i^2=v_f^2\\\Rightarrow -v_e^2+v_i^2=v_f^2\\\Rightarrow v_f=√(v_i^2-v_e^2)](https://img.qammunity.org/2020/formulas/physics/college/3su9wm0aovjwirfqpzh3wuznubedc0fc96.png)
= Escape velocity of Earth = 11.2 km/s
= Velocity of projectile = 44.4 km/s
![v_f=√(44.4^2-11.2^2)\\\Rightarrow v_f=42.96\ km/s](https://img.qammunity.org/2020/formulas/physics/college/ksvzlsuyydkfqn9mguf8o286y3ir8m370h.png)
The velocity of the spacecraft when it is more than halfway to the star is 42.96 km/s