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A body is projected vertically from earth's surface with velocity equal to half the escape velocity. The maximum height reached by the satellite is ( $R=$ radius of earth)
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The correct answer is:
$\frac{\mathrm{R}}{3}$
Given: $\mathrm{v}=\frac{\mathrm{v}_{\mathrm{c}}}{2}$
If body is projected with velocity $\mathrm{v}\left(\mathrm{v} < \mathrm{v}_e\right)$ then height up to which it will rise, $h=\frac{R}{\left(\frac{v_c^2}{v^2}-1\right)}$
$\therefore \quad \mathrm{h}=\frac{\mathrm{R}}{\left(\frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{e}} / 2}\right)^2-1}=\frac{\mathrm{R}}{4-1}=\frac{\mathrm{R}}{3}$
If body is projected with velocity $\mathrm{v}\left(\mathrm{v} < \mathrm{v}_e\right)$ then height up to which it will rise, $h=\frac{R}{\left(\frac{v_c^2}{v^2}-1\right)}$
$\therefore \quad \mathrm{h}=\frac{\mathrm{R}}{\left(\frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{e}} / 2}\right)^2-1}=\frac{\mathrm{R}}{4-1}=\frac{\mathrm{R}}{3}$
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