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. `u costheta`
B. `(u)/(2)sqrt(1+cos^(2)theta)`
C. `(u)/(2)sqrt(1+2cos^(2)theta)`
D. `(u)/(2)sqrt(1+3cos^(2)theta)`

Answer

4

Explanation

`v=("displacement")/("time")=(sqrt(((R)/(2))^(2)+H^(2)))/(T//2)`
`=sqrt(((u_(0)^(2)sin2theta)/(2g))^(2)+((u_(0)^(2)sin^(2)theta)/(2g))^(2))/((u_(0)sintheta)/(g))`
`=((u_(0)^(2)sin2theta)/(2g)sqrt(4cos^(2)theta+sin^(2)theta))/((u_(0)sintheta)/(g))`
`=(u_(0))/(2)sqrt(1+3cos^(2)theta)`