In case of freely falling body, the ratio of KE of the body at the end of 4th second to increase in KE in the next four seconds is
1:1
1:2
1:3
1:4
KE4ΔKE=12mv4212mv82−v42=(g×4)2(g×8)2−(g×4)2=g2×16(64−16)g2=1648=13