Explanation

(b) Refer figure are when projectile is at $A$, then
[image:4_image136.emf]
$OC = \frac{R}{2} = \frac{1}{2}\frac{u^{2}}{g}\sin{2\theta} = \frac{1}{2} \times \frac{\left( 20\sqrt{2} \right)^{2}}{10}\sin 2 \times 45{^\circ}$
$= \ 40\ m$
$AC = H = \frac{u^{2}\sin^{2}\theta}{2g} = \frac{\left( 20\sqrt{2} \right)^{2}}{2 \times 10}\sin^{2}{45{^\circ}}$
$\therefore$ Displacement, $OA = \sqrt{OC^{2} + CA^{2}} = \sqrt{40^{2} + 20^{2}}$
Time of projectile from $O$ to $A$
$= \frac{1}{2}\left( \frac{2u\sin\theta}{g} \right) = \frac{u\sin\theta}{2g} = \frac{\left( 20\sqrt{2} \right)^{\ }\sin{45{^\circ}}}{10} = 2s$
$\therefore\ Average\ velocity\ = \frac{displacement\ }{time}$
$= \frac{\sqrt{40^{2} + 20^{2}}}{2} = 10\sqrt{5}\ ms^{- 1}$