A mass falls from a height ' h ' and its time of fall ' t ' is...

The distance covered by the mass falling from height ' $h$ ' during its time of fall ' $t$ ' is given by $$\begin{array}{r}\mathrm{s}=\mathrm{h}=\mathrm{ut}+\frac{1}{2} \mathrm{gt}^2 \\ \text{ As, } \mathrm{u}=0 \Rightarrow \mathrm{h}=\frac{1}{2} \mathrm{gt}^2 \Rightarrow \mathrm{t}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}} \end{array}$$The time period of simple pendulum is $\mathrm{T}=$2 $\pi \sqrt{\frac{\mathrm{l}}{\mathrm{g}}}$where, lis the length of the pendulum. From Eq. (i) and (ii), since ' $h$ ' and ' $l$ ' are constant so, we can conclude that, $\mathrm{t} \propto \frac{\mathrm{l}}{\sqrt{\mathrm{g}}} \text{ and } \mathrm{T} \propto \frac{\mathrm{l}}{\sqrt{\mathrm{g}}} \quad \therefore \frac{\mathrm{t}}{\mathrm{T}}=\mathrm{l}$Thus, the ratio of time of fall and time period of pendulum is independent of value of gravity $(\mathrm{g})$ or any other parameter like mass and radius of the planet. Thus, the relation between $\mathrmt^{\prime}$ and $\mathrm{T}^{\prime}$ on another planet irrespective of its mass or radius will remains same as it was on earth i.e. \mathrmt^{\prime}=2 $\mathrm{~T}^{\prime}$