Correct Answer - C
We know that escape velocity, `v_(e)=sqrt((2GM_(e))/(R_(e)))`. Substituting the values, we get `v_(e) = 112 " kms"^(-1)`. But if the projected velocity is double then the speed become `sqrt(3)` times and having the value `112 xx sqrt(3)=19.4 " kms"^(-1)`.