Given that, initial velocity, $u=5 \mathrm{~m} / \mathrm{s}$
and horizontal range $=2 \times$ greatest height attained or
$\begin{array}{ll}
\text { or } & H=2 \times R \Rightarrow \frac{u^2 \sin ^2 \theta}{2 g}=\frac{u^2 \sin 2 \theta}{g} \\
\Rightarrow & \frac{\sin ^2 \theta}{2}=\sin 2 \theta \Rightarrow \frac{\sin ^2 \theta}{2}=2 \sin \theta \cos \theta \\
\Rightarrow & \frac{\sin \theta}{2}=2 \cos \theta \Rightarrow \frac{\sin \theta}{\cos \theta}=4
\end{array}$
$\Rightarrow \quad \tan \theta=4 \Rightarrow \tan \theta=\frac{4}{1}=\frac{\text { Perpendicular }}{\text { Base }}$
If we form a right angled triangle, with these sides, then


Now, $\sin \theta=\frac{4}{\sqrt{17}}$ and $\cos \theta=\frac{1}{\sqrt{17}}$
So, the horizontal range
$\begin{aligned}
R & =\frac{u^2 \sin 2 \theta}{g}=\frac{u^2 \times 2 \sin \theta \cos \theta}{g} \\
& =\frac{(5)^2 \times 2 \times \frac{4}{\sqrt{17}} \times \frac{1}{\sqrt{17}}}{10}=\frac{25 \times 2 \times 4}{10 \times 17}=\frac{20}{17} \\
& =1.17 \mathrm{~m} \approx 2 \mathrm{~m}
\end{aligned}$