A particle is projected at time t=0 from a point P wilth a speed `v_0` at an angle of `45^@` to the horizontal. Find the magnitude and the direction of the angular momentum of the parallel about the point P at time `t=v_0/g`

Explanation

Let us take the origin at P, X-axis along the horizontal and Y-axis along the vertically upward direction as shown in ure. For horizontal motion during the time 0 to t`,

`v_x=v_0cos45^@=v_0/sqrt2`
and `x=v_xt=v_0/sqrt2.v_0/g=v_0^2/(sqrt2 g)`
for vertical motion, ltbr. `v_y=v_0sin45^@-gt=v_0/sqrt2-v_0=((1-sqrt2))/sqrt2v_0`
and `y=(v_0sin45^@)t-1/2gt^2`
`v_0^2(sqrt2g)-v_0^2/(2g)=v_0^2/(2g)(sqrt2-1)`
the angular momentum of the particle at time t about the origin is
`L=vecrxxvecp=mvecrxxvecv`
`=m(vecix+vecjy)xx(veciv_x+vecjv_y)`
`=m(veckxv_y-veckyv_x)`
`=mveck[(v_0^2/(sqrt2g)v_0/sqrt2(1-sqrt2)-v_0^2/(2g)(sqrt2-1)v_0/sqrt2]`
`-veck(mv_0^3)/(sqrt2 sg)`
Thus the angular momentum of the particle is `(mv_0^2)/(2sqrt2g)` in the negative Z direcrtion i.e. perpendicuar to the plane of motion going into the plane.