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# A body is projected vertically upwards from the surface of the earth with a velocity equal to $\large\frac{3}{4}$th escape velocity of earth. If 'R' is the radius of earth, the maximum height attained by the body is :

$\begin {array} {1 1} (1)\;\large\frac{9}{7}R & \quad (2)\;\large\frac{9}{8}R \\ (3)\;\large\frac{10}{3}R & \quad (4)\;\large\frac{10}{9}R \end {array}$

none of the above

because with 3/4 of escape velocity the object cannot reach the height R how can it go above R

(1) $\large\frac{9}{7}R$
Explanation:
According to the law of conservation of energy,
$(T.E)_{at\;surface} = (T.E)_{at\;height}$
$(K.E+P.E)_{surface} =(K.E+P.E)_{max \;height}$
$\large\frac{1}{2}mv^2 +(\large\frac{-GMm}{R})$ =$0+(\large\frac{-GMm}{R+h})$
$Given V= \large\frac{3}{4}v_e=\large\frac{3}{4}\sqrt{\large\frac{2GM}{R}}$
$=\large\frac{1}{2}m \times \large\frac{9}{16} \times (\large\frac{2GM}{R})+(\large\frac{-GMm}{R})$
$=(\large\frac{-GMm}{R+h})$
$\large\frac{9GMm}{16R}-\large\frac{GMm}{R}=-\large\frac{GMm}{R+h}$
$\large\frac{-7GMm}{16R}=-\large\frac{GMm}{R+h}$
$\large\frac{7}{16R}=\large\frac{1}{R+h}$
$\Rightarrow 7R+7h=16R$
$h=\large\frac{9}{7}R$