Average torque on a projectile of mass m, initial speed u and angl,e of projection `theta` between initial and final positions P and Q as shown in figure about the point of projectiion is

Options

A. `(mu^(2)sin2theta)/2`
B. `mu^(2)costheta`
C. `mu^(2)sintheta`
D. `(mu^(2)costheta)/2`

Answer

A

Explanation

`vectau_(av).,/_\t=/_\vecL`
`|/_\vecL|=|vecL_(f)=vecL_(i)|` above point of projecton
`=(musintheta)("Range")=0`
`=((musintheta)(u^(2)sin2theta))/g=(mu^(3)sinthetasin2theta)/g`
`|vectau_(av)|=(vec/_\L)/(/_\t)=(mu^(3)sinthetasin2theta)/g g/(2usintheta)`
`(mu^(2)sintheta)/2`