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7.Gravitation
hard
An object is projected vertically up from the earth's surface with velocity $\sqrt {Rg}$ where $R$ is the radius of the earth and $'g'$ is the acceleration due to earth on the surface of earth. Find the maximum height reached by the object.
A
$\frac{R}{2}$
B
$R$
C
$2R$
D
$3R$
Solution

Let the object of mass $\mathrm{m}$ projected with speed
$\mathrm{u}=\sqrt{\mathrm{Rg}}$ reach a height $'h'$ above surface of
earth.
Then from conservation of energy
$\frac{1}{2} m u^{2}-\frac{G M m}{R}=-\frac{G M m}{R+h}$
$\mathrm{u}^{2}=\mathrm{g} \mathrm{R}=\frac{\mathrm{GM}}{\mathrm{R}}$
$\Rightarrow \frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{R}}-\frac{\mathrm{GMm}}{\mathrm{R}}=-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}$ or $\mathrm{h}=\mathrm{R}$
Standard 11
Physics