A ball of mass m is thrown at an angle `alpha` to the horizontal with the initial velocity `v_0`. Find the time dependence of the magnitude of the ball's angular momentum vector relative to the point from which the ball is thrown. Find the angular momentum M at the highest point of the trajectory if `m=130g`, `alpha=45^@`, and `v_0=25m//s`. The air drag is to be neglected.

Explanation

`vecM(t)=vecrxxvecp=(vecv_0t+1/2vecg t^2)xxm(vecv_0+vecg t)`
`=mv_0g t^2sin (pi/2+alpha)(-veck)+1/2mv_0g t^2sin (pi/2+alpha)(veck)`
`=1/2mv_0g t^2cos alpha(-veck)`:
Thus `M(t)=(mv_0g t^2cos alpha)/(2)`
Thus angular momentum at maximum height
i.e. at `t=tau/2=(v_0sin alpha)/(g)`
`M(pi/2)=((mv_0^3)/(2g))sin^2alphacosalpha=37kg-m^2//s`
Alternate:
`vecM(0)=0` so, `vecM(t)=underset(0)overset(t)intvecNdt=underset(0)overset(t)int(vecrxxmvecg)`
`=underset(0)overset(t)int[(vecv_0t+1/2vecg t^2)xxmvecg]dt=(vecv_0xxmvecg)t^2/2`