\[(a)\;\frac{\sqrt {u_1}}{g}\quad (b)\;\frac{\sqrt {u_2}}{g} \quad (c)\;\frac{\sqrt {u_1u_2}}{g} \quad (d)\;\frac{2 \sqrt {u_1u_2}}{g}\]

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If $\overrightarrow {S_1} \perp \overrightarrow{S_2}$

The displacement vectors are perpendicular

Then $\overrightarrow {S_1} . \overrightarrow{S_2}=0$

$\overrightarrow {S_1}=u_1t \hat i-\large\frac{1}{2}$$gt^2 \hat j$

$\overrightarrow {S_2}=-u_2t \hat i-\large\frac{1}{2}$$gt^2 \hat j$

$\overrightarrow {S_1} . \overrightarrow{S_2}=0$

=>$-u_1u_2t^2+\large\frac{1}{4}$$g^2t^4=0$

=>$4u_1u_2t^2=g^2t^4$

=>$\large\frac{4u_1u_2}{g^2}$$=t^2$

=>$t=\large\frac{2 \sqrt {u_1u_2}}{g}$

Hence d is the correct answer.

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