A, B, C and D are points in a vertical line such that AB = BC = CD. If a body falls from rest at A, then the times of descent through ,AB, BC and CD are in the ratio

Let ,AB = BC = CD = h. Suppose time taken in covering the distances AB, BC and CD be t₁ , t₂ and t₃ respectively. Then
For AB part, $\mathrm{h}=\frac{1}{2}{\mathrm{gt}}_1^2$
For AC part, $2\mathrm{h}=\frac{1}{2}\mathrm{g}{\left({\mathrm{t}}_1+{\mathrm{t}}_2\right)}^2$
For AD part, $3\mathrm{h}=\frac{1}{2}\mathrm{g}{\left({\mathrm{t}}_1+{\mathrm{t}}_2+{\mathrm{t}}_3\right)}^2$
$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{t}}_1:\left({\mathrm{t}}_1+{\mathrm{t}}_2\right):\left({\mathrm{t}}_1+{\mathrm{t}}_2+{\mathrm{t}}_3\right)=1:\sqrt{2}:\sqrt{3}\\ \mathrm{or} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{t}}_1:{\mathrm{t}}_2:{\mathrm{t}}_3=1:(\sqrt{2}-1):(\sqrt{3}-\sqrt{2})\end{array}$