Conserving mechanical energy at the surface of earth and the maximum height attained,
$\frac{-G M m}{R_e}+\frac{1}{2} m \frac{4 G M}{3 R_e^2} R_e=\frac{-G M m}{r}+0$
$P_{. E_i}+K \cdot E_i=P_{.} E_f+K_i E_f$
$\Rightarrow \frac{-G M m}{R_e}+\frac{2 G M m}{3 R^e}=\frac{-G M m}{r}$
$-\frac{1}{3} \frac{G M m}{R_e}=\frac{-G M m}{r}$
$\Rightarrow r=3 R_e$
$\Rightarrow R_e+h=3 R_e$
$h=2 R_e$
Now, let us calculate the velocity of the particle at height equal to half of the maximum height $i . e$ at $h=R_e$ Again using mechanical conservation of energy,
$P . E_i+K . E_i=P . E_j+K . E_j$
$\frac{-G M m}{R_e}+\frac{1}{2} m \frac{4}{3} \frac{G M}{R_e^2} \times R_e=\frac{-G M m}{2 R}+\frac{1}{2} m v^2$
$\Rightarrow -\frac{1}{3} \frac{G M m}{R_e}+\frac{G M m}{2 R_e}=\frac{1}{2} m v^2$
$\Rightarrow \frac{G M m}{6 R_e}=\frac{1}{2} m v^2$
$\Rightarrow v=\sqrt{\frac{G M}{3 R_e}}=\sqrt{\frac{G M}{R_e^2} \times \frac{R_e}{3}}=\sqrt{\frac{g R_e}{3}}$
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