A particle is projected at time $t = 0$ from a point $P$ on the ground with a speed $v_0,$ at an angle of ${45}^{\circ }$ to the horizontal. The magnitude of angular momentum of the particle about $P$ at time $t = v_0/g$  is

 

$At time t =v_0/g,$

$Let L= angular momentum of the particle about P,$

$P_x= x-component of momentum of particle,$

$v_x = x-component of velocity of particle and$

$x = displacement along x-axis.$

$Let p_y , v_y, and y denote the quantities along y-axis$

$Angular momentum, L = x p_y - y p_x$

 $or L = x \left(m{v}_y\right) - y\left(m{v}_x\right) or L = m\left[x v_y - y v_x\right] \hat{k} \dots \left(i\right)$

$To find v_x and v_y :$

$v_x = Horizontal velocity of the projectile$

$or v_x = v_0 \cos {45}^{\circ } = \frac{v_0}{\sqrt{2}} \dots \left(ii\right)$

It remains constant all along.

      $v_y = vertical velocity of particle$

$or v_y = \left(v_0 \sin {45}^{\circ }\right) - g \times \left(\frac{v_0}{g}\right) \left[\because v = u + at\right]$

$or v_y = \left(\frac{v_0}{\sqrt{2}}- v_0\right) \dots \dots \left(iii\right)$

$To find x and y :$

$or x = \frac{v_0}{\sqrt{2}} \times \frac{v_0}{g} = \frac{v_0^2}{\sqrt{2}g} \therefore x =\frac{v_0^2}{\sqrt{2}g} \dots .\left(iv\right)$

$y = v_0 \sin \theta t - \frac{g{t}^2}{2}$

$y = \frac{v_0}{\sqrt{2}} \times \frac{v_0}{g} - \frac{g}{2}{\left(\frac{v_0}{g}\right)}^2$

$or y = \left(\frac{v_0^2}{\sqrt{2}g}- \frac{v_0^2}{2g}\right) \dots .\left(v\right)$

$\therefore Substitute \left(ii\right), \left(iii\right), \left(iv\right) and \left(v\right) in \left(i\right)$

       $L = m \left[\frac{v_0^2}{\sqrt{2}g}\times \left(\frac{v_0}{\sqrt{2}}- v_0\right) - \left(\frac{v_0^2}{\sqrt{2}}-\frac{v_0^2}{2g} \right) \frac{v_0}{\sqrt{2}}\right]$

$or L = \frac{m{v}_0^3}{g} \left[ \left(\frac{1}{2} - \frac{1}{\sqrt{2}}\right) - \left(\frac{1}{2} - \frac{1}{2\sqrt{2}}\right) \right]$

$or L = \frac{m{v}_0^3}{g} \times \left(\frac{-1}{2\sqrt{2}}\right) = \frac{-m{v}_0^3}{2\sqrt{2}g}$

$L$ is $\perp$ to plane of motion and is directed away from the reader i.e., along $-ve z direction.$