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

Options

A. `v/2sqrt(1+2 cos^(2) theta)`
B. `v/2sqrt(1-4 cos^(2) theta)`
C. `v/2sqrt(1+3 cos^(2) theta)`
D. `v cos theta`

Answer

C

Explanation

Average velocity =`("Displacement")/("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)sin2 theta)/g`
and T=time of flight =`(2v sin theta)/g`
`v_(av)=v/2sqrt(1+3cos^(2) theta)`