• # question_answer A particle is projected under gravity with velocity $\sqrt{2ag}$ from a point at a height h above the level plane at an angle $\theta$ to it. The maximum range R on the ground is           A) $\sqrt{\left( {{a}^{2}}+1 \right)h}$            B)        $\sqrt{{{a}^{2}}h}$                C) $\sqrt{ah}$                D)        $2\sqrt{a(a+h)}$

[d] Coordinates of point P are $(R,-h)$. Hence $-h=R\tan \theta -\frac{g{{R}^{2}}}{2\left( 2ga \right)}\left( 1+{{\tan }^{2}}\theta \right)$ $\Rightarrow \,\,\,\,{{R}^{2}}{{\tan }^{2}}\theta -4aR\,\,\tan \theta +({{R}^{2}}-4ah)=0$ For $\theta$ to be real, ${{(4aR)}^{2}}\ge 4{{R}^{2}}({{R}^{2}}-4ah)$ $\Rightarrow \,\,\,4{{a}^{2}}\ge ({{R}^{2}}-4ah)\Rightarrow {{R}^{2}}\le 4a(a+h)$ $\Rightarrow \,\,R\le 2\sqrt{a(a+h)}\,\,\,\,\,\Rightarrow \,\,\,\,\,\,{{R}_{\max }}=2\sqrt{a(a+h)}$