A lift of total mass, M, is raised by cables from rest to 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 be made in shortest time is :

Weight of lift $=Mg$
Maximum tension $=nMg$
$\therefore$ Maximum acceleration $=(\mathrm{nMg}-\mathrm{Mg})/\mathrm{M}=(\mathrm{n}-1)\mathrm{g}$
And maximum retardation $=g$
A corresponding velocity-time graph for shortest time will be as shown:

Here $(n-1)g=v_m/t_1$............(1)
Or ${\mathrm{t}}_1={\mathrm{v}}_{\mathrm{m}}/(\mathrm{n}-1)\mathrm{g}$
And $g=v_m/t_2$.............(2)
Or ${\mathrm{t}}_2={\mathrm{v}}_{\mathrm{m}}/\mathrm{g}$
Area under v -t graph is total displacement h .
Hence, $h=1/2(t_1+t_2)v_m$
From above equations we get,
${\mathrm{V}}_{\mathrm{m}}=\sqrt{2\mathrm{gh}\left(\frac{\mathrm{n}-\mathrm{l}}{\mathrm{n}}\right)}$