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Two particles move in a uniform gravitational field with an acceleration g. At the initial moment the particles were located at one point and moved with velocities u_(1)=3.0ms^(-1) and u_(2)=4.0ms^(-1) horizontally in opposite directions. Find the distance betwen the particles at the moment when their velocity vectors become mutually perpendicular.

Answer» <html><body><p></p>Solution :The situation is shown fig. <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AI_PHY_XI_V01_A_C05_SLV_022_S01.png" width="80%"/> <br/> Let the velocity vectors become <a href="https://interviewquestions.tuteehub.com/tag/perpendicular-598789" style="font-weight:bold;" target="_blank" title="Click to know more about PERPENDICULAR">PERPENDICULAR</a> after time t when both particles has fallen same vertical distance `1/2"<a href="https://interviewquestions.tuteehub.com/tag/gt-1013864" style="font-weight:bold;" target="_blank" title="Click to know more about GT">GT</a>"^(2)` and have acquired same vertical velocites gt. <br/> Let their <a href="https://interviewquestions.tuteehub.com/tag/resultant-1187362" style="font-weight:bold;" target="_blank" title="Click to know more about RESULTANT">RESULTANT</a> velocities make angle `theta_(1)` and `theta_(2)` with horizontal. <br/> Then `tan theta_(1)=("gt")/(v_(1))` and `tan theta_(2)=("gt")/(v_(2))` <br/> Since velocity vectors are perpendicular <br/> `theta_(1)+theta_(2)=90^(@)`, hence `tan theta_(2)=cot theta_(1)` <br/> It makes `tan theta_(1)=("gt")/(v_(1))` and `cot theta_(1)=("gt")/(v_(2))` <br/> On multiplying we <a href="https://interviewquestions.tuteehub.com/tag/get-11812" style="font-weight:bold;" target="_blank" title="Click to know more about GET">GET</a> `t=(sqrt(v_(1)v_(2)))/g` and `D=(u_(1)+u_(2))t`</body></html>


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