A freely falling body moves from A to B during  2sec and from B to C during 1sec. If AB=BC the distance between A&B is

Let ‘O’ be the starting point of the body and let ‘t’ be the time taken by it to move from O to A .Given that from A to B it takes 2sec And from B to C it takes  1secAB=BC⇒(OB−OA)=OC−OB⇒12g(t+2)2−t2=12g(t+3)2−(t+2)2⇒4t+t=2t+5⇒2t=1⇒t=12s.The distance between A&B is12g(t+2)2−t2=12g[4t+4]=12g[2+4]=6g2=3g