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),

\(R_{max}=4R_{max}\)