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2.Motion in Straight Line
hard
Two bodies begin a free fall from the same height at a time interval of $N s$. If vertical separation between the two bodies is $1$ after $n\, second$ from the start of the first body, then $n$ is equal to
A
$\sqrt {nN} $
B
$\frac{1}{{gN}}$
C
$\frac{1}{{gN}} + \frac{N}{2}$
D
$\frac{1}{{gN}} - \frac{N}{4}$
Solution
$\begin{array}{l}
{y_1} = \frac{1}{2}g{n^2},\,{y_2} = \frac{1}{2}{\left( {n – N} \right)^2}\\
\therefore \,\,{y_1} – {y_2} = \frac{1}{2}g\left[ {{n^2} – {{\left( {n – N} \right)}^2}} \right]\\
\Rightarrow \,1 = \frac{g}{2}\left( {2n – N} \right)N\\
\left[ {\,{y_1} – {y_2} = 1} \right]\\
\Rightarrow \,n = \frac{1}{{gN}} + \frac{N}{2}
\end{array}$
Standard 11
Physics
