A particle is projected from the ground with an initial speed u at an angle `theta` with the horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is

Explanation

Average velocity ` ("displacmemnt")/("time")`
`V_(av) = (sqrt(H^(2) +(R^(2))/(4)))/((T)/(2))" "……..(i)`
Here H = maximum height ` = (v^(2) sin^(2) theta)/(2g)`
`R = "range" = (v^(2) sin 2theta)/(g)` and T = time of flight `= (2 v sin theta)/(g)`
Substituting in (i) , we get .
`V_(av) = (v)/(2) sqrt(1+ 3cos^(2) theta)`