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))`

Options

A. Its velocity after `sqrt(2)` sec will be perpendicular to its initial velocity
B. Its velocity after `sqrt(3)` sec will be perpendicular to its initial velocity
C. Its radius of curvature is 40m at `x=5(sqrt(3)+1)` m
D. Its radius of curvature is 5m at `x=5m`

Explanation

`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`