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  4. A stone falls freely such that the distance covered by It in the last second of its motion is equal to the distance covered by it in the first 5 s. It is in air for:

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Q. A stone falls freely such that the distance covered by It in the last second of its motion is equal to the distance covered by it in the first $5\, s$. It is in air for:

Jharkhand CECEJharkhand CECE 2004

Solution:

In case of freely falling body initial velocity is zero.
From equation of motion the distance travelled in the $n$th second of motion is
$s_{n}=u+\frac{1}{2} g(2 n-1)$ ....(i)
Also $s=u t+\frac{1}{2} g t^{2}$ .....(ii)
Putting $u=0$ in Eqs. (i) and (ii)
and $t=5 s$ in Eq. (ii), we get
$s=\frac{1}{2} g(5)^{2}=\frac{25}{2} g$,
$ s_{n}=\frac{1}{2} g(2 n-1)$
Given, $s_{n}=s $
$\therefore \frac{1}{2} g(2 n-1)=\frac{25}{2} \times g $
$\Rightarrow 2 n-1=25$
$ \Rightarrow n=13 s$