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A particle is projected with velocity 10m/s at an angle of `45^(@)` with the horizontal from some height. Consider point of projection as origin and motion in xy plane. Then `(vecg=10hatj m//s^(2))`A. Its velocity after `sqrt(2)` sec will be perpendicular to its initial velocityB. Its velocity after `sqrt(3)` sec will be perpendicular to its initial velocityC. Its radius of curvature is 40m at `x=5(sqrt(3)+1)` mD. Its radius of curvature is 5m at `x=5m` |
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Answer» `vecv_(1)=(10)/sqrt(2)hati+(10)/sqrt(2)hatj` after `sqrt(2)` sec. `vec_(2)=(10)/sqrt(2)hati((10)/sqrt(2)-10sqrt(2))hatj` particle will beat x=5 after `t=(1)/sqrt(2)` sec `rArrv_(y)=0` R.O.C. `=(v_(bot^(2))/(a_(bot)))rArr((10)/sqrt(2))^(2)/(10)rArr5M` |
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