A particle is projected with a velocity \( v \) so that its horizontal range twice the greatest height
attained. The horizontal range is

Velocity of particle \(=v\)
Horizontal range \(=2 \times\) Greatest height attained
Now, Range \(=\frac{v^{2} \sin 2 \theta}{g}\)
Height \(=\frac{v^{2} \sin ^{2} \theta}{2 g}\)
\(\Rightarrow \frac{v^{2} \sin 2 \theta}{g}=\frac{2 \times v^{2} \sin ^{2} \theta}{2 g} \Rightarrow \sin 2 \theta=\sin ^{2} \theta\)
Using, \(\sin 2 \theta=2 \sin \theta \cos \theta\), we get
\(2 \sin \theta \cos \theta=\sin ^{2} \theta \Rightarrow 2 \cos \theta=\sin \theta \Rightarrow \frac{\sin \theta}{\cos \theta}=2\)
\(\Rightarrow\) tan \(\theta=2\)
\(\Rightarrow \sin \theta=\frac{2}{\sqrt{5}}\) and \(\cos \theta=\frac{1}{\sqrt{5}}\)
Therefore,
Range
\(=\frac{v^{2} \sin 2 \theta}{g}=\frac{2 v^{2} \sin \theta \cos \theta}{g}=\frac{2 v^{2}}{g} \frac{2}{\sqrt{5}} \times \frac{1}{\sqrt{5}}=\frac{4}{5} \frac{v^{2}}{g}\)
\(\Rightarrow\) Horizontal range \(=\frac{4}{5} \frac{v^{2}}{g}\)