Question

Two bodies begin a free fall from the same height at a time interval of $N$ seconds. If vertical separation between the two bodies is $1$ m after n seconds 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. $\frac{1}{gN}-\frac{N}{4}$

Answer 1

The correct option is C $\frac{1}{gN}+\frac{N}{2}$

Let $x_1$ be the distance travelled by the first body.

Let $x_2$ be the distance travelled by the second body.

Using $s=ut+\frac{1}{2}g{t}^2$,

$\Rightarrow -x_1=0\times n-\frac{1}{2}g{n}^2$ $\Rightarrow x_1=\frac{1}{2}g{n}^2$,

$\Rightarrow -x_2=0\times (n-N)-\frac{1}{2}g(n-N{)}^2$ $\Rightarrow x_2=\frac{1}{2}g(n-N{)}^2$,

Given $x_1-x_2=1=\frac{1}{2}g{n}^2-\frac{1}{2}g(n-N{)}^2$ $\Rightarrow n=\frac{1}{gN}+\frac{N}{2}$