Two particles are thrown horizontally in opposite direction with velocities $u$ and $2 u$ from the top of a high tower. The time after which their radius of curvature will be mutually perpendicular is:

\sqrt{2} \frac{u}{g}

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2 \frac{u}{g}

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\frac{1}{\sqrt{2}} \frac{u}{g}

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\frac{1}{2} \frac{u}{g}

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Solution

The correct option is A $\sqrt{2} \frac{u}{g}$
$_{1} = u (-^i) + (- g t)^j$
${\to v}_{2} = 2 u (^i) + (- g t)^j$
$\Rightarrow_{1} ._{2} = 0$
$\Rightarrow - 2 u^{2} + g^{2} t^{2} = 0$
$\Rightarrow t = \sqrt{\frac{2 u^{2}}{g^{2}}} = \sqrt{2} . \frac{u}{g}$
The radius of curvature will be mutually perpendicular only when the velocity vectors will be mutually perpendicular i.e. after the time $t = \sqrt{2} \frac{u}{g}$ .

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Two particles are thrown horizontally in opposite direction with velocities u and 2u from the top of a high tower. The time after which their radius of curvature will be mutually perpendicular is:

The correct option is A √2ug→v1=u(−^i)+(−gt)^j→v2=2u(^i)+(−gt)^j⇒→v1.→v2=0⇒−2u2+g2t2=0⇒t=√2u2g2=√2.ugThe radius of curvature will be mutually perpendicular only when the velocity vectors will be mutually perpendicular i.e. after the time t=√2ug.

Two particles are thrown horizontally in opposite direction with velocities $u$ and $2 u$ from the top of a high tower. The time after which their radius of curvature will be mutually perpendicular is: