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A rocket of mass $M$ is launched vertically from the surface of the earth with an initial speed $V.$ Assuming the radius of the earth to be $R$ and negligible air resistance, the maximum height attained by the rocket above the surface of the earth is
$\frac{R}{{\left( {\frac{{gR}}{{2{V^2}}} - 1} \right)}}$
${R\left( {\frac{{gR}}{{2{V^2}}} - 1} \right)}$
$\frac{R}{{\left( {\frac{{2gR}}{{{V^2}}} - 1} \right)}}$
${R\left( {\frac{{2gR}}{{{V^2}}} - 1} \right)}$
Solution
$\Delta \mathrm{K} \cdot \mathrm{E}=\Delta \mathrm{U}$
$\Rightarrow \frac{1}{2} \mathrm{MV}^{2}=\mathrm{GM}_{\mathrm{e}} \mathrm{M}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}+\mathrm{h}}\right)$ $…(i)$
Also $\mathrm{g}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}^{2}}$ $…(ii)$
On solving ( $i$ ) and $(ii)$ $\mathrm{h}=\frac{\mathrm{R}}{\left(\frac{2 \mathrm{g} \mathrm{R}}{\mathrm{V}^{2}}-1\right)}$
