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 = Mg
Maximum tension=nMg

$\therefore \text{ Maximum acceleration }=\frac{nMg-Mg}{M}$

                                                $=(n-1)g$

and maximum retardation = g
corresponding velocity-time graph for shortest time will be as follows:

$\text{ Here }(n-1)g=\frac{v_m}{t_1}\text{ or }{t}_1=\frac{v_m}{(n-1)g} .......\left(1\right)$

$\text{ and }g=\frac{v_m}{t_2}\text{ or }{t}_2=\frac{v_m}{g} ......\left(2\right)$

Area under v-t graph is total displacement ft.

$\mathrm{Hence} h=\frac{1}{2}\left(t_1+t_2\right)v_m ......\left(3\right)$

From (1), (2) and (3) we get,

$v_m=\sqrt{2gh\left(\frac{n-1}{n}\right)}$