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(B) Let $M$ be the mass of the earth and $m$ be the mass of the body. The escape velocity is given by $V_e = \sqrt{2GM/R}$.The initial velocity of the

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$R/3$

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(B) Let $M$ be the mass of the earth and $m$ be the mass of the body. The escape velocity is given by $V_e = \sqrt{2GM/R}$.The initial velocity of the body is $v = V_e/2 = \frac{1}{2} \sqrt{2GM/R}$.Using the law of conservation of energy between the surface of the earth and the maximum height $h$:$K_i + U_i = K_f + U_f$$\frac{1}{2}mv^2 - \frac{GMm}{R} = 0 - \frac{GMm}{R+h}$Substituting $v^2 = \frac{1}{4} V_e^2 = \frac{1}{4} \left( \frac{2GM}{R} \right) = \frac{GM}{2R}$:$\frac{1}{2}m \left( \frac{GM}{2R} \right) - \frac{GMm}{R} = - \frac{GMm}{R+h}$$\frac{GMm}{4R} - \frac{GMm}{R} = - \frac{GMm}{R+h}$$-\frac{3GMm}{4R} = - \frac{GMm}{R+h}$$\frac{3}{4R} = \frac{1}{R+h}$$3(R+h) = 4R$$3R + 3h = 4R$$3h = R$$h = R/3$

Column-$I$	Column-$II$
$1. m = -2$	$a$. Convex mirror
$2. m = -\frac{1}{2}$	$b$. Concave mirror
$3. m = +2$	$c$. Real Image
$4. m = +\frac{1}{2}$	$d$. Virtual image

$A$ body is projected vertically from the surface of the earth of radius $R$ with a velocity equal to half of the escape velocity. The maximum height reached by the body is

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