A body is projected vertically upwards from the surface of earth...

Applying energy conservation from ($\left(1\right)\text{to}\left(2\right)$,
$\dfrac{1}{2}m \cdot \left(\dfrac{2GM}{{R}_{e}}\right) - \dfrac{GMm}{{R}_{e}} = \dfrac{1}{2}m{v}^{2} - \dfrac{GMm}{R + r}$

$⇒ \dfrac{1}{2}m{v}^{2} = \dfrac{GMm}{R + r}$

$⇒ v = \sqrt{\dfrac{2GM}{R + r}} = \dfrac{dr}{dt}$

$⇒ \sqrt{2GM}{∫}_{0}^{t}dt = {∫}_{{R}_{e}}^{{R}_{e} + h} ( \sqrt{R + r} ) dr$

$\sqrt{2GM} \cdot t = \dfrac{2}{3}{\left[( R + r) \dfrac{3}{2}\right]}_{{R}_{e}}^{{R}_{e} + h}$

$t = \dfrac{2}{3}\sqrt{{R}_{e}^{3}2GM}\left[\left(1 + \dfrac{h}{{R}_{e}}\right)\dfrac{3}{2} - 1\right]$

$GM{R}_{e}^{2} = g$

$t = \dfrac{1}{3}\sqrt{\dfrac{2{R}_{e}}{g}}\left[\left(1 + \dfrac{h}{{R}_{e}}\right)\dfrac{3}{2} - 1\right]$