Ask Questions, Get Answers

Want to ask us a question? Click here
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11
0 votes

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

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

Can you answer this question?

1 Answer

0 votes
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$
$\overrightarrow {S_1} . \overrightarrow{S_2}=0$
=>$t=\large\frac{2 \sqrt {u_1u_2}}{g}$
Hence d is the correct answer.
answered Jun 28, 2013 by meena.p
edited May 24, 2014 by lmohan717

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App