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Two particles are projected horizontally from the same elevated point in opposite directions with velocities $4 m s^{- 1}$ and $9 m s^{- 1}$ respectively. At the moment when their velocity vectors are mutually perpendicular, the separation between them is (g=10ms^-2)

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5 m at angle ${tan}^{- 1} \frac{9}{4}$

3 m at angle ${tan}^{- 1} \frac{9}{4}$

7.8 m horizontally

5m horizontally

answer is C.

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Detailed Solution

$S= (u_{1} {+u}_{2}) \frac{\sqrt{u_{1} u_{2}}}{g} = \frac{13×2×3}{10} =7.8m$

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