The horizontal distance between two bodies, when their velocity are perpendicular to each other, is

1 m

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0.5 m

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2 m

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4 m

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Solution

The correct option is C $4 m$
Velocity of first body at any instant $t, {\to v}_{1} = 2^i - g t^j$
Velocity of second body at any instant
${\to v}_{2} = - 8^i - g t^j$
Since, ${\to v}_{1} ⊥ {\to v}_{2}$ , i.e., ${\to v}_{1} \cdot {\to v}_{2} = 0$
$(2^i - g t^j) \cdot (- 8^i - g t^j) = 0 \Rightarrow - 16 + g^{2} t^{2} = 0$
$\Rightarrow t = \sqrt{\frac{16}{g^{2}}} \Rightarrow t = \frac{4}{10} = 0.4 s$
${\to S}_{1} = (2 \times 0.4)^i, {\to S}_{2} = (8 \times 0.4) (-^i)$
${\to S}_{1} = 0.8^i$ and ${\to S}_{2} = - 3.2^i$
Separation $= {\to S}_{1} - (- {\to S}_{2}) = 0.8^i + (3.2^i) = 4^i$
${\to S}_{1} = 2 t^i - \frac{1}{2} g t^{2}^i, {\to S}_{2} = - 8 t^i - \frac{1}{2} g t^{2}^j$
As ${\to S}_{1} ⊥ {\to S}_{2}$ i.e., ${\to S}_{1} \cdot {\to S}_{2} = 0$
$- 16 t + \frac{1}{2} \times \frac{1}{2} g^{2} t^{4} = 0 \Rightarrow g^{2} t^{2} = 16 \times 4$
$\Rightarrow g t = 4 \times 2 \Rightarrow t = \frac{8}{10} = 0.8 s$ .

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