On a planet,a freely falling body takes 2 , s | vedclass

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(A) For a freely falling body dropped from rest $(u = 0)$ from a height $h$,the distance covered is given by $h = \frac{1}{2} g_P t^2$,where $g_P$ is

Vedclass · Accepted Answer

$3.14$

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(A) For a freely falling body dropped from rest $(u = 0)$ from a height $h$,the distance covered is given by $h = \frac{1}{2} g_P t^2$,where $g_P$ is the acceleration due to gravity on the planet.Given $h = 8 \, m$ and $t = 2 \, s$,we have $8 = \frac{1}{2} 	imes g_P 	imes (2)^2$.$8 = 2 	imes g_P \implies g_P = 4 \, m/s^2$.The time period $T$ of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g_P}}$.Substituting $l = 1 \, m$ and $g_P = 4 \, m/s^2$,we get $T = 2\pi \sqrt{\frac{1}{4}} = 2\pi 	imes \frac{1}{2} = \pi \, s$.Since $\pi \approx 3.14$,the time period is $3.14 \, s$.

On a planet,a freely falling body takes $2 \, s$ when it is dropped from a height of $8 \, m$. The time period of a simple pendulum of length $1 \, m$ on that planet is ..... $s$.

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