Two bodies start falling freely from the rest from the same height at an interval of 1 s. How long after the first body begins to fall will the two bodies be 10 m apart? (g = 10 $m / s^{2}$ )

0.5 s

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1.0 s

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1.5 s

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2.0 s

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Solution

The correct option is C 1.5 s
Let $x_{1}$ be the distance travelled by first body.
Let $x_{2}$ be the distance travelled by second body.
Let after $n$ seconds after fall of first body , both bodies to be 10 $m$ apart.
Using $s = u t + \frac{1}{2} g t^{2}$ ,
$\Rightarrow - x_{1} = 0 \times n - \frac{1}{2} g n^{2}$ $\Rightarrow x_{1} = \frac{1}{2} g n^{2}$ ,
$\Rightarrow - x_{2} = 0 \times (n - 1) - \frac{1}{2} g (n - 1)^{2}$ $\Rightarrow x_{2} = \frac{1}{2} g (n - 1)^{2}$ ,
Given $x_{1} - x_{2} = 10 = \frac{1}{2} g n^{2} - \frac{1}{2} g (n - 1)^{2}$ $\Rightarrow n = 1.5$ $s$

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