Question

A particle is projected with a velocity $u$ such that its range on horizontal plane is twice the greatest height attained by it. The range of the projectile will be
$(g$ is the acceleration due to gravity).

• A. $\frac{4u^2}{5g}$
• B. $\frac{4u^2}{\sqrt{5g}}$
• C. $\frac{u^2}{g}$
• D. $\sqrt{\frac{u}{g}}$

Answer 1

The correct option is A $\frac{4u^2}{5g}$

$\text{Step 1: Range and height of projectile [Refer Fig. 1]}$

Range of the projectile, $R=\frac{u^2\sin 2\theta }{g}$

Maximum height, $H=\frac{u^2{\sin }^2\theta }{2g}$

$R=2H$ (given)

$\Rightarrow \frac{u^2\sin 2\theta }{g}=$$\frac{2\times u^2{\sin }^2\theta }{2g}$

$\Rightarrow \frac{2\times u^2\sin \theta cos\theta }{g}=$$\frac{2\times u^2{\sin }^2\theta }{2g}$

$\Rightarrow$ $\cot \theta =\frac{1}{2}$

$\text{Step 2: Calculations [Ref. Fig. 2]}$

From the triangle

$\sin \theta =\frac{2}{\sqrt{5}}$ and $\cos \theta =\frac{1}{\sqrt{5}}$

Range of the projectile

$R=\frac{2u^2\sin \theta \cos \theta }{g}=\frac{2u^2}{g}\frac{2}{\sqrt{5}}\times \frac{1}{\sqrt{5}}$

$=\frac{4u^2}{5g}$

Hence option $A$ is correct.