A body is projected horizontally from the surface of the Earth with a velocity equal to m times the escape velocity. Neglect rotational effects of the earth. The maximum height attained by the body from the earth's surface is R/2 . Then n must be : 

\(\sqrt{0.6}\)

\(nM v_e R = mv(R + \frac{R}{2})\)

\(\Rightarrow v = \frac{2}{3}n v_e \text{; } v_e = \sqrt{\frac{2GM}{R}} \)

\(\frac{GMm}{R} - \frac{GMm}{3R/2} = \frac{1}{2} m (nv_e)^2 - \frac{1}{2} mv^2\)

\(\Rightarrow \frac{1}{3} \frac{GMm}{R} = \frac{1}{2} m n^2 v^2_e (1 - \frac{4}{9})\)

\(\Rightarrow \frac{1}{3} \frac{GMm}{R} = \frac{1}{2} m n^2 \frac{2GM}{R}  \frac{5}{9}\)

\(n = \sqrt{\frac{3}{5}} = \sqrt{0.6}\)