1.

At a given instant of time the position vector a particle moving in a circle with a velocity 3 hati - 4 hatj +5k is hati + 9 hatj -3 hatk. Itsangular velocity at that time is

Answer»

`(( 13 hati+ 29 hatj -31 hatk))/sqrt146`
`(( 13 hati- 29 hatj -31 hatk))/146`
`(( 13 hati+ 29 hatj -31 hatk))/sqrt146`
`(( 13 hati+ 29 hatj +31 hatk))/sqrt146`

SOLUTION :MAXIMUM height ,
`H=1/2g t^2=1/2g((t_1+t_2)/2)^2`
(`:.` The TIME taken by the body to REACH maximum height,where time of ascending is equal to time of Decending ,`t=(t_1+t_2)/(2)`
`:.H=2g(t_1+t_2)/(4^2)`


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