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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11
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Two stones are projected horizontally from top of a tower in opposite directions with velocities $u_1$ and $u_2$. The time after which their displacement vectors are perpendicular to each other is

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

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1 Answer

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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.
answered Jun 28, 2013 by meena.p
edited May 24, 2014 by lmohan717
 

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