A lift of total mass M is raised by cables from rest through a height h. The greatest tension which the cables can safely bear is nMg. The maximum speed of lift during its journey if the ascent is to made in shortest time is

Weight of lift = MgMaximum tension=nMg∴  Maximum acceleration =nMg−MgM                                                =(n−1)gand maximum retardation = gcorresponding velocity-time graph for shortest time will be as follows: Here (n−1)g=vmt1 or t1=vm(n−1)g    .......(1) and g=vmt2 or t2=vmg    ......(2)Area under v-t graph is total displacement ft.Hence h=12t1+t2vm  ......(3)From (1), (2) and (3) we get,vm=2ghn−1n