A mass falls from a height ‘h’ and its time of fall ‘t’ is recorded in terms of time period T of a simple pendulum. On the surface of earth it is found that t = 2T. The entire set up is taken on the surface of another planet whose mass is half of that of earth and radius the same. Same experiment is repeated and corresponding times noted as t' and T'. Then we can say

Step 1: Given

Height = $h$

time = $t$

Time period of simple pendulum = $T$

On Earth's surface, $t=2T$

Mass of the other planet, $M_p=\frac{1}{2}{M}_e$

acceleration due to gravity, $g$

Step 2: Formula used and calculation

Time period of simple pendulum

$T=2\pi \sqrt{\frac{l}{g}}$

$g=\frac{G{M}_e}{R^2}$

where, $R$ is radius of earth

Now,

using $s=ut+\frac{1}{2}a{t}^2$

$h=0+\frac{1}{2}a{t}^2$

$h=\frac{1}{2}g{t}^2$

$t=\sqrt{\frac{2h}{g}}=2T$

$\frac{2h}{g}=4T^2$

$h=2g{T}^2$

On surface of other planet

$g_p=\frac{G{M}_p}{R^2}$

$g_p=\frac{G\frac{M_e}{2}}{R^2}$

$g_p=\frac{g}{2}$

$T^{\prime }=2\pi \sqrt{\frac{l}{g_p}}$

$T^{\prime }=2\pi \sqrt{\frac{2l}{g}}$

$T^{\prime }=\sqrt{2}T$

Again using

$s=ut+\frac{1}{2}a{t}^2$

$h=\frac{1}{2}{g}_p{t}^{\prime 2}$

${t^{\prime }}^2=\frac{2h}{g_p}$

$t^{\prime }=\sqrt{\frac{2h\times 2}{g}}$

$t^{\prime }=2\sqrt{\frac{2T^{2}g}{g}}$

$t^{\prime }=2\sqrt{2}T$

$t^{\prime }=2\sqrt{2}\frac{T^{\prime }}{\sqrt{2}}$

$t^{\prime }=2T^{\prime }$