InterviewSolution
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InterviewSolution
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A particle is projected with a velocity `vecv=8hati+6hatj m//s`. The time after which it will starts moving perpendicular to its initial direction of motions isA. `0.5s`B. `1.25s`C. `1s`D. `5//3s` |
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Answer» Correct Answer - d Given `vecv_(1)=8hati+6hatjm//s^(2)` Acceleration due to gravity, `veca=-ghati=-10hatjm//s^(2)` Velocity after time t, `vecv_(2)=vecv_(1)+vec(a)t` `vecv_(2)=(8hati+6hatj)-10thatj=8hatj+(6-10t)hatj` Now `vecv_(1) and vecv_(2)` will be perpendicular if `vecv_(1).vecv_(2)=0` `(8hati+6hatj).[8hati+(6-10t)hatj]=0` `64+6(6-10t)=0` `64+36-60t=0` `60t=100t=5/3 sec` |
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