Explanation

(d) The minimum velocity of projection to achieve escape velocity can be calculated as
$\ \ \ \ \ \ \ \ \ \ \ \ \ Intial\ KE = \frac{1}{2}{mv}^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2} \times m(4 \times 11.2)^{2} = 16 \times \frac{1}{2}{mv}_{e}^{2}$
$As\frac{1}{2}{mv}_{e}^{2}$ energy is used up in coming out from the gravitational pull of the earth, so
final KE should be$\ 15 \times \frac{1}{2}{mv}_{e}^{2}$
$Hence,\ \ \ \ \ \ \ \frac{1}{2}{mv}_{e}^{2} = 15 \times \frac{1}{2}{mv}_{e}^{2}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\ v}^{'2} = {15v}_{e}^{2}$
$or\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v^{'} = \sqrt{15}v_{e}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \sqrt{15} \times 11.2{kms}^{- 1}$