A particle is projected at time t = 0 from a point P on the ground with a speed `v_0`, at an angle of `45^(@)` to the horizontal. The angular momentum of the particle about P at time t = `v_0`/g is

Explanation

`(mv_(0)^(3)//(2sqrt(2)g)`
Let us take the origin at P, x-axis along the horizontal and y-axis along the vertically upward direction as shown in the following figure.
For horizontal motion during the time 0 to t,

`v_(x) = v_(0) cos 45^(@) = v_(0)/sqrt(2)` and `x=v_(x)t = v_(0)/sqrt(2) v_(0)/g = v_(0)^(2)/sqrt(2)`

`y=v_(0)/sqrt(2) (v_(0)/g)-1/2 g(v_(0)/g)^(2) = v_(0)^(2)/(2g) (sqrt(2)-1)`
The angular momentum of the particle at time t about the origin is
`L= vecr xx vecp = m(vecr xx vecv) = (veci x + vecjy) xx (v_(x) veci + v_(y)vecj) =m(kxv_(y)-kyv_(x))`
`=mveck [(v_(0)^(2)/(sqrt(2)g) v_(0)/sqrt(2)(1-sqrt(2))-v_(0)^(2)/(2g)(sqrt(2)-1)v_(0)/(sqrt(2)))]=-veck (mv_(0)^(3))/(2sqrt(2)g)`
Thus, the angular momentum of the particles is `mv_(0)^(3)//(2sqrt(2)g)` in the negative Z-direction, i.e., perpendicular to the plane of motion, going into the plane.