An object of mass 5kg is thrown vertically upwards from the ground....

Net force while going upward will be $mg + R = 50 + 10 = 60N$

$⇒ {a}_{up} = \dfrac{60}{5} = 12m{s}^{- 2}$

Similarly, Net force while going downward will be $mg - R = 40N$

$⇒ {a}_{d} = \dfrac{40}{5} = 8m{s}^{- 2}$

Since

$h = \dfrac{1}{2}a{t}^{2} ⇒ \dfrac{{a}_{up}{{t}_{up}}^{2}}{{a}_{d}{{t}_{d}}^{2}} = 1 ⇒ \dfrac{{t}_{up}}{{t}_{d}} = \sqrt{\dfrac{8}{12}} = \sqrt{\dfrac{2}{3}}$