A particle is released from rest from a tower of height 3h. The ratio of times to fall equal heights h, i.e.,${\text{t}}_{\text{1}}{\text{:t}}_{\text{2}}{\text{:t}}_{\text{3}}$is 

$$\begin{gathered} \text{h}=\frac{1}{2}{{\text{gt}}_1}^2 \\ {\text{T}}_1=\sqrt{\frac{\text{2h}}{\text{g}}}=\sqrt{2}\sqrt{\frac{h}{g}} \\ \text{2h}=\frac{1}{2}{{\text{gt}}_2}^2 \\ {\text{t}}_{\text{2}}={\text{T}}_{\text{2}}-{\text{T}}_{\text{1}} \\ {\text{T}}_{\text{2}}=2\sqrt{\frac{\text{h}}{\text{g}}} \\ =2\sqrt{\frac{\text{h}}{\text{g}}}-\sqrt{2}\sqrt{\frac{\text{h}}{\text{g}}} \\ =(2-\sqrt{2})\sqrt{\frac{\text{h}}{\text{g}}} \\ \text{3h}=\frac{1}{2}{{\text{gt}}_3}^2 \\ {\text{T}}_3=\sqrt{\frac{\text{6h}}{\text{g}}} \\ {\text{t}}_{\text{3}}={\text{T}}_{\text{3}}-{\text{T}}_{\text{2}} \\ =\sqrt{6}\sqrt{\frac{\text{h}}{\text{g}}}-2\sqrt{\frac{\text{h}}{\text{g}}} \\ =(\sqrt{6}-2)\sqrt{\frac{\text{h}}{\text{g}}} \\ \frac{\frac{t_1}{t_2}}{t_3}=\frac{\frac{\sqrt{2}\sqrt{\frac{h}{g}}}{(2-\sqrt{2})\sqrt{\frac{h}{g}}}}{(\sqrt{6}-2)\sqrt{\frac{h}{g}}} \\ =\frac{\frac{1}{(\sqrt{2}-1)}}{(\sqrt{3}-\sqrt{2})} \end{gathered}$$