As object return with smaller speed that means there is a loss of energy. This loss of energy occurs due to viscous drag of atmosphere. While object is moving up let viscous drag produces an deceleration ' a '. Then, ‌v2−u2=2as ⇒‌‌0−(10)2=2(−(g+a))h....(i) ‌ or ‌‌‌100=2(g+a)h While coming down,
‌⇒‌‌64−0=2(−(g−a))(−h) ‌⇒‌‌64=2(g−a)h.....(ii) Dividing (i) and (ii) we get ‌‌
100
64
=‌
g+a
g−a
⇒‌‌100g−100a=64g+64a ‌36g=164a‌ or ‌a=‌
36
164
g Substituting ' a ' in either (i) or (ii) we get ' h. From (ii) ‌64‌=2(g−‌