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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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Two bodies begin to free fall from same height at a time interval of N seconds, If the vertical seperation between them is 1m after n seconds from the start of 1st body, then n equals

\[(a)\;\sqrt {nN} \quad (b)\;\frac{1}{gN} \quad (c)\;\frac{1}{gN}+\frac{N}{2}\quad (d)\frac{1}{gN}-\frac{N}{4}\]

Can you answer this question?
 
 

1 Answer

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$s_1=\large\frac{1}{2}$$gn^2$
$s_2=\large\frac{1}{2}$$g(n-N)^2$
$s_1-s_2=\large \frac{1}{2}$$ g[n^2-(n-N)^2]$
$1=\large\frac{g}{2}$$[2n-N]N$
$n=\large\frac{1}{gN}+\frac{N}{2}$
Hence c is the correct answer.

 

answered Jun 24, 2013 by meena.p
edited Jan 25, 2014 by meena.p
 

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