Question

A particle is released from rest from a tower of height 3h.The ratio of times to fall equal height h,i.e., t${}_1$: t${}_2$ :t${}_3$ is

• A. $\sqrt{3}:\sqrt{2}$: 1
• B. 3 : 2 : 1
• C. 9 :4 : 1
• D. 1 : ($\sqrt{2}$ - 1) :($\sqrt{3}-\sqrt{2}-1$)

Answer 1

The correct option is D 1 : ($\sqrt{2}$ - 1) :($\sqrt{3}-\sqrt{2}-1$)

Time in falling first height $h$ will be $t_1=T_1=\sqrt[2]{2\frac{2h}{g}}$

Time in falling first height $2h$ will be $T_2=\sqrt[2]{2\frac{2\left(2h\right)}{g}}$

so the time in falling second $h$ will be $t_2=T_2-T_1=\sqrt[2]{2\frac{h}{g}}(\sqrt[2]{24}-\sqrt[2]{22})$

Time in falling in height $3h$ will be $T_3=\sqrt[2]{2\frac{2\left(3h\right)}{g}}$

so the time in falling third $h$ will be $t_3=T_3-(T_1+T_2)=\sqrt[2]{2\frac{h}{g}}(\sqrt[2]{26}-\sqrt[2]{24}-\sqrt[2]{22})$

the ratio will be $t_1:t_2:t_3=\sqrt[2]{22}:\sqrt[2]{22}(\sqrt[2]{22}-1):\sqrt[2]{22}(\sqrt[2]{23}-\sqrt[2]{22}-1)$

$\sqrt[2]{22}$ is common in all so can be cancelled so the answer is option D.