Question

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. $\sqrt{(a^2+1)h}$
• B. $\sqrt{a^{2}h}$
• C. $\sqrt{ah}$
• D. $2\sqrt{a(a+h)}$

Answer 1

Answer: (D)

$2\sqrt{a(a+h)}$

$\text{Coordinates of point P are}(R,-h)$

$\text{Hence,}h=R\tan \theta -\frac{g{R}^2}{2\left(2ga\right)}(1+{\tan }^2\theta )\Rightarrow R^2{\tan }^2\theta -4aR\tan \theta +(R^2-4ah)=0$

$\text{For}\theta \text{to be real,}(4aR{)}^2\ge 4R^2(R^2-4ah)$

$\Rightarrow 4a^2\ge (R^2-4ah)\Rightarrow R^2\le 4a(a+h)\Rightarrow R\le 2\sqrt{a(a+h)}\Rightarrow R_{max}=2\sqrt{a(a+h)}$