A particle is prejected from ground with speed 80 m/s at an angle `30^(@)` with horizontal from ground. The magnitude of average velocity of particle in time interval t = 2 s to t = 6 s is [ take g = 10 m/`s^(2)` ]

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`.