A ball is dropped from height h=100 m,from surface of planet . if In the last  12s of its journey, it covers a distance of 19 m.Then acceleration due to gravity (in ms-2)  on that planet is -----------

Time taken to fall on ground=t=2ha=2100a=200a.....1 Time taken to cover initial distance of (100-19 =81m) t'=2h'a ⇒t−12=281a.....2  dividing (1) by (2) tt−12=10081 ⇒tt−12=109 ⇒9t=10t-5 ⇒t=5 Substituting in (1) and then squaring it ⇒25=200a ⇒a=8