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) sin 2theta)/2`
B. `m u^(2) cos theta`
C. `m u^(2) sin theta`
D. `(m u^(2) cos theta)/2`

Answer

A

Explanation

`tau_(av).Deltavect = DeltaL` Here, `Deltat`= time of flight `=(2mu sin theta)/g`……….(1)
`|DeltavecL| =|(vecL_(f) - vecL_(j))|` about point of projection.
`=(mu sin theta)("range") -0 =((m u sintheta)(mu^(2) sin 2theta))/g`
`(mu u^(2) sintheta 2theta)/g therefore |vectau_(av)| =(vecDeltaL)/(Deltat) = (mu^(3) sin theta sin 2theta)/g .g/(2u sin theta) = (m u^(2) sin 2theta)/2`