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At a certain height a body at rest explodes into two equal fragments with one fragment receiving a horizontal velocity of 10 ms^-1. The horizontal distance between the two fragments when their position vectors are perpendicular to each other is

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40m

10m

20m

answer is A.

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Detailed Solution

${\bar{S}}_{1} = ut \hat{i} - \frac{1}{2} {gt}^{2} \hat{j}$

${\bar{S}}_{2} = - ut \hat{i} - \frac{1}{2} {gt}^{2} \hat{j}$

${\bar{S}}_{1}$ perpendicular to ${\bar{S}}_{2}$ then

${\bar{S}}_{1} . {\bar{S}}_{2} = 0$

$t = \frac{2 u}{g} = \frac{2 \times 10}{10} = 2 s$

Distance between them when displacement vectors are $⊥$ r is
d = (u₁ + u₂)t = 2ut = 2(10)(2) = 40m

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