Maximum height reached by a rocket fired with a speed equal to 50% of the escape velocity from earth's surface is:

V = $\frac{50}{100}{\mathrm{V}}_{\mathrm{e}} = \frac{1}{2}\sqrt{\frac{2\mathrm{GM}}{\mathrm{R}}}$

Apply energy conservation

$\Rightarrow -\frac{\mathrm{GMm}}{\mathrm{R}}+\frac{1}{2}{\mathrm{mV}}^2 = -\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})}$

         ${\mathrm{v}}^2 = \frac{2\mathrm{GM}}{\mathrm{R}}-\frac{2\mathrm{GM}}{\mathrm{R}+\mathrm{h}}$

          $\frac{1}{4}.\frac{2\mathrm{GM}}{\mathrm{R}} = 2\mathrm{GM}(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}+\mathrm{h}})$

           $\frac{1}{4\mathrm{R}}= \frac{\mathrm{h}}{\mathrm{R}(\mathrm{R}+\mathrm{h})}$

          R+h = 4h $\Rightarrow \mathrm{h} = \frac{\mathrm{R}}{3}$