A particle of mass 100g is projected at time t = 0 with a speed 20m...

As torque due to gravitational force will be variable, therefore we can use

$\DeltaL = {∫}_{0}^{t}τdt$.

Now, torque of $mg$ at time $t = \left({v}_{x}t\right)mg$ and horizontal velocity of the projectile will remain constant\left({v}_{x} = vcos45 ° = 20 \times \dfrac{1}{\sqrt{2}} = 10\sqrt{2}\right)$.</p><p>${L}_{0} = {∫}_{0}^{2}mg\left({v}_{x}t\right)dt$</p><p>$= mg{v}_{x}\dfrac{{t}^{2}}{2} = ( 0 . 1 ) ( 10 ) ( 10\sqrt{2} ) \dfrac{{2}^{2}}{2}$</p><p>$= 20\sqrt{2}$</p><p>$= \sqrt{800}kg{m}^{2}{s}^{- 1}$</p><p>Therefore,$K = 800.