Prove that the maximum horizontal range is four times the maximum heig

1.	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.
Answer» 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), \(R_{max}=4R_{max}\)

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.

Answer»

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),

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