The horizontal range is maximum for θ = 45º and it is given by
\(R_{max}=\frac{u^2}{g}\)
The maximum height attained,
\(H=\frac{u^2sin^2\,\theta}{2g}\) ...(i)
Therefore, for θ = 45º
\(H_{max}=\frac{u^2sin^2\,45^o}{2g}\) ...(ii)
From the equation (i) and (ii),
\(R_{max}=4R_{max}\)