A body is thrown from the surface of the earth with velocity 'V' \(\frac{\mathrm{m}}{\mathrm{s}}\). The maximum height above the earth's surface upto which it will reach is \((\mathrm{R}=\) radius of earth, \(\mathrm{g}=\) acceleration due to gravity)

\(\begin{aligned} \frac{1}{2} \mathrm{mu}^{2} &=-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}-\left(-\frac{\mathrm{GMm}}{\mathrm{R}}\right)=\frac{\mathrm{GMm}}{\mathrm{R}}-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}} \\ &=\mathrm{GMm}\left[\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}+\mathrm{h}}\right] \\ \therefore \mathrm{u}^{2} &=2 \mathrm{GM}\left[\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}+\mathrm{h}}\right] \\ \mathrm{u}^{2} &=2 \mathrm{gR}^{2}\left[\frac{\mathrm{R}+\mathrm{h}-\mathrm{R}}{\mathrm{R}(\mathrm{R}+\mathrm{h})}\right] \\ \mathrm{u}^{2} &=2 \mathrm{gR}\left[\frac{\mathrm{h}}{\mathrm{R}+\mathrm{h}}\right] \\ \therefore \quad \frac{\mathrm{u}^{2}}{2 \mathrm{gR}} &=\left[\frac{\mathrm{h}}{\mathrm{R}+\mathrm{h}}\right] \quad \therefore \quad \frac{\mathrm{R}+\mathrm{h}}{\mathrm{R}}=\frac{2 \mathrm{gR}}{\mathrm{u}^{2}} \\ \frac{\mathrm{R}}{\mathrm{h}}+1 &=\frac{2 \mathrm{gR}}{\mathrm{u}^{2}} \\ \therefore \frac{\mathrm{R}}{\mathrm{h}} &=\frac{2 \mathrm{gR}}{\mathrm{u}^{2}}-1=\frac{2 \mathrm{gR}-\mathrm{u}^{2}}{\mathrm{u}^{2}} \therefore \mathrm{h}=\frac{\mathrm{Ru}^{2}}{2 \mathrm{gR}-\mathrm{u}^{2}} \end{aligned}\)