A lift on the ground starts from rest and accelerates uniformly upward at 1.25m/s² , reaching its maximum velocity in 10s . When it has attained a height of 40m, a bolt gets loosened and falls to the ground. What is the height of the lift (in m) when the bolt hits the ground?

$\begin{array}{l}\mathrm{S}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2\\ Distance covered by lift in 10 sec:\\ \mathrm{S}=0+\frac{1}{2}(1.25)(10{)}^2\\ \Rightarrow \mathrm{S}=62.5\mathrm{m}\\ \Rightarrow \mathrm{V}=0+1.25\times 10\\ \Rightarrow \mathrm{V}=12.5\mathrm{m}/\mathrm{s}\\ time taken to cover 40 m distance: \\ 40=\frac{1}{2}(1.25){\mathrm{t}}^2\\ \Rightarrow \mathrm{t}=8\mathrm{s}\\ velocity at 40 m distance :\\ \mathrm{V}=0+1.25\times 8\\ \Rightarrow \mathrm{V}=10 \mathrm{m}/\mathrm{s}\end{array}$
For BOLT,

$\mathrm{S}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$
$\begin{array}{l}-40=10\mathrm{t}-5{\mathrm{t}}^2\\ \Rightarrow {\mathrm{t}}^2-2\mathrm{t}-8=0\\ \Rightarrow \mathrm{t}=4\mathrm{s}\end{array}$

Time taken by bolt to reach the ground = 4 s
Distance travelled by lift from t =8s to t =10s
$\begin{array}{l}{\mathrm{S}}_1=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2\\ =10\times 2+\frac{1}{2}\times 1.25\times 2^2=22.5 m\\ distance\mathit{ }travelled\mathit{ }in\mathit{ }t\mathit{=}\mathit{10}\mathit{ }s\mathit{ }to\mathit{ }t\mathit{=}12 s\\ {\mathrm{S}}_2=12.5\times 2=25\mathrm{m}\end{array}$
hence, Height of lift = 40 + 22.5 + 25 = 87.5m