Average torque on a projectile of mass m (initial speed u and angle of projection $\mathrm{\theta }$) between initial and final positions P and Q as shown in figure, about the point of projection is:

$\overrightarrow{{\mathrm{\tau }}_{\mathrm{av}}}.∆\mathrm{t} = ∆\mathrm{L} ------\left(\mathrm{i}\right)$

Here $∆\mathrm{t} = \mathrm{time} \mathrm{of} \mathrm{flight} = \frac{2\mathrm{u} \sin \mathrm{\theta }}{\mathrm{g}}$

Change in angular momentum about point of projection (initially it is zero)

$\left|\overrightarrow{∆\mathrm{L}}\right| = |{\overrightarrow{\mathrm{L}}}_{\mathrm{f}}-\overrightarrow{{\mathrm{L}}_{\mathrm{i}}}| = (\mathrm{mu} \sin \mathrm{\theta }) \mathrm{Range}$

         = $\frac{(\mathrm{mu} \sin \mathrm{\theta })({\mathrm{u}}^2 \sin 2\mathrm{\theta })}{\mathrm{g}} = \frac{{\mathrm{mu}}^3\sin \mathrm{\theta } \sin 2\mathrm{\theta }}{\mathrm{g}}$

Now $\left|{\overrightarrow{\mathrm{\tau }}}_{\mathrm{av}}\right| = \left|\frac{\overrightarrow{∆\mathrm{L}}}{∆\mathrm{t}}\right| = \frac{{\mathrm{mu}}^3\sin \mathrm{\theta } \sin 2\mathrm{\theta }}{\mathrm{g}}\times \frac{\mathrm{g}}{2\mathrm{u} \sin \mathrm{\theta }}$

         = $\frac{{\mathrm{mu}}^2 \sin 2\mathrm{\theta }}{2}$