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Prove that the maximum horizontal range is four times the maximum height attained by the projectile, when fired at an inclination so as to have maximum horizontal range.

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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),

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

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