A rocket starts vertically upwards with speed $v_0$. Choose the correct option(s) (R is the radius of the earth).

Kinetic energy on the surface of earth = $\frac{1}{2}\mathrm{m}{v}_0^2$

Potential energy on the surface of earth = $-\frac{\mathrm{GMm}}{\mathrm{R}}$

Total energy $=\frac{1}{2}\mathrm{m}{v}_0^2-\frac{\mathrm{GMm}}{\mathrm{R}}$

Kinetic energy at a height $\mathrm{h}=\frac{1}{2}\mathrm{m}{v}^2$

Potential energy at this height $=\frac{-\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})}$

Total energy $=\frac{1}{2}\mathrm{m}{v}^2-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}$

By the principle of conservation of energy,

$\frac{1}{2}\mathrm{m}{v}^2-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}=\frac{1}{2}\mathrm{m}{v}_0^2-\frac{\mathrm{GMm}}{\mathrm{R}}$

$\frac{1}{2}(v_0^2-v^2)=\frac{\mathrm{GM}}{\mathrm{R}}-\frac{\mathrm{GM}}{\mathrm{R}+\mathrm{h}}$

But $\mathrm{GM}={\mathrm{gR}}^2$

$\therefore \frac{1}{2}(v_0^2-v^2)=\frac{{\mathrm{gR}}^2\mathrm{h}}{\mathrm{R}(\mathrm{R}+\mathrm{h})}$

$v_0^2-v^2=\frac{2\mathrm{gRh}}{\mathrm{R}+\mathrm{h}}=\frac{2\mathrm{gh}}{1+\frac{\mathrm{h}}{\mathrm{R}}}$

At maximum height $v=0.$

The initial velocity $v_0=\left(90\mathrm{\%}\right)$

$v_{\text{escape}}=0.9\sqrt{2\mathrm{gR}}$  then

$(0.9\sqrt{2\mathrm{gR}}{)}^2-0=\frac{2{\mathrm{gh}}_{max}}{1+\frac{{\mathrm{h}}_{max}}{\mathrm{R}}};0.81\mathrm{R}=\frac{{\mathrm{h}}_{max}}{1+\frac{{\mathrm{h}}_{max}}{\mathrm{R}}}$

$0.81\mathrm{R}+0.81{\mathrm{h}}_{max}={\mathrm{h}}_{max};0.19{\mathrm{h}}_{max}=0.81\mathrm{R}$

Maximum height reached by the rocket

${\mathrm{h}}_{max}=\frac{0.81\mathrm{R}}{0.19}=4.26\mathrm{R}$