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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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The distance through which a body falls in the nth seconds is h. The distance through which it falls in the next seconds is

$\begin{array}{1 1} h \\ h+\large\frac{g}{2} \\ h-g \\ h+g \end{array} $

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Answer : $ h+g$
$y= \large\frac{1}{2}$$g(n+1)^2-\large\frac{1}{2} $$gn^2$
$\quad= \large\frac{R}{2} $$[ (n+1)^2 -n^2]$
$\quad= \large\frac{R}{2} $$(2n+1)$
Also, $h= \large\frac{g}{2} $$(2n-1)$
$\therefore y= \large\frac{g}{2} \bigg[ \large\frac{2h}{g}$$ +1+1 \bigg]$
or $y= h+g$
answered Aug 19, 2014 by meena.p

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