A particle is project from ground with a speed of `6 m//s` at an angle of `cos^(-1)(sqrt((7)/(27)))` with horizontal. Then the magnitude of average velocity of particle from its starting point to the highest point of its trajectory in `m//s` is:

Answer

`4`

Explanation

Average velocity from `A` to `B`
`v_(avg) = (AB)/(T//2) = sqrt((R//2)^(2) + H^(2))/(T//2)`
`v_(avg) = (u)/(2) sqrt(1 + 3cos^(2)theta)`
`= (6)/(2)sqrt(1 + 3 xx (7)/(27) ) = 4 m//s`.