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Two particles are simultaneously projected in opposite directions horizontally from a given point in space, where gravity g is uniform. If $u_{1} and u_{2}$ and be their initial speeds, then the time t after which their velocities are mutually perpendicular is given by

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$\frac{\sqrt{u_{1}^{2} + u_{2}^{2}}}{g}$

$\frac{\sqrt{u_{1} u_{2}}}{g}$

$\frac{\sqrt{u_{2} (u_{1} + u_{2})}}{g}$

$\frac{\sqrt{u_{1} (u_{1} + u_{2})}}{g}$

answer is A.

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Detailed Solution

Given, $v_{1} ⊥ v_{2}$

$∴ v_{1} \cdot v_{2} = 0$

$or (u_{1} \hat{i} - g t \hat{j}) \cdot (- u_{2} \hat{i} - g t \hat{j}) = 0$

$∴ g^{2} t^{2} = u_{1} u_{2} or t = \frac{\sqrt{u_{1} u_{2}}}{g}$

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