A particle of mass `m` is projected at `t=0` from the point `P` on the ground with speed `v_(0)` at an angle of `theta` to the horizontal. Find the magnitude and direction of the angular momentum of the particle at time `t=(v_(0)sintheta)/(2g)`.

Explanation

`t=(v_(0)sintheta)/(2g)`
`x:v_(x)=v_(0)costheta`
`x=v_(0)costheta t=v_(0) costhetaxx(v_(0)sintheta)/(2g)=(v_(0)^(2)sintheta costheta)/(2g)`
`y:v_(y)=v_(0)sintheta-"gt"`
`=v_(0)sintheta-gxx(v_(0)sintheta)/(2g)=(v_(0)sintheta)/(2)`
`y=v_(0) sintheta t-(1)/(2)"gt"^(2)`
`=v_(0)sinthetaxx(v_(0)sintheta)/(2g)-(1)/(2)g((v_(0)sintheta)/(2g))^(2)`
`=(v_(0)^(2)sin^(2)theta)/(2g)-(v_(0)^(2)sin^(2)theta)/(8g)=(3)/(8)(v_(0)^(2)sin^(2)theta)/(g)`
`L=mv_(x)y-mv_(y)x`
`=m(v_(0)costhetaxx(3)/(8)(v_(0)^(2)sin^(2)theta)/(g)-(v_(0)sintheta)/(2)xx(v_(0)^(2)sinthetacostheta)/(2g))`
`=(nv_(0)^(3)sin^(2)thetacostheta)/(8g)`
OR
In vector form
`vecv=v_(x)hati+v_(y)hatj,vecr=xhati+yhatj`
`vecL=vecrxxvecp=vecrxxmvecv=m|(i,j,k),(x,y,o),(v_(x),v_(y),o)|`
`=m(xy_(y-yv_(x)))hatk`
`=m[(v_(0)^(2)sinthetacostheta)/(2)xx(v_(0)sintheta)/(2)-(3)/(8)xx(v_(0)^(2)sin^(2)theta)/(g)xxv_(0)costheta]hatk`
`m=[-(1)/(8)(v_(0)^(3)sin^(2)thetacostheta)/(g)]hatk^(~)`
`vecL= (-mv_(0)^(3)sin^(2)thetacostheta)/(8g)hatk`
`L=|vecL|=(mv_(0)^(3)sin^(2)thetacostheta)/(8g)`