What is the maximum height attained by a body projected with a velocity equal to one-third of the escape velocity from the surface of the earth? (Radius of the earth = R)                               [AIIMS 2017]

SHORT CUT :    WHEN   V=$\frac{V_E}{N} then the height reached by the body h=\frac{R}{N^2-1}$

 From conservation of energy,

$\frac{1}{2}m{v}_1^2-\frac{G{M}_{e}m}{R}=\frac{1}{2}m{v}_2^2-\frac{G{M}_{e}m}{R+H}$

Here,v₂ = 0                                   (At maximum height)

$\therefore \frac{1}{2}m{v}_1^2-\frac{G{M}_{e}m}{R}=-\frac{G{M}_{e}m}{R+H}$

Solving, we get

$H=\frac{R}{\frac{2gR}{v_1^2}-1}=\frac{R}{\frac{v_e^2}{v_1^2}-1} \left(\because v_e=\sqrt{2gR}\right)$

where, v_e is escape velocity.

Here,  $v_1=\frac{1}{3}{v}_e\Rightarrow \frac{v_e}{v_1}=3$

$\therefore H=\frac{R}{9-1}=\frac{R}{8}$