Question

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

Answer 1

Formula for Height and Range:

$\boxed{\bf H_{max} = \dfrac{u^2 sin^2\theta}{2g}}$

$\boxed{\bf R = \dfrac{u^2 sin(2\theta)}{g}}$

• where "u" is initial velocity, "g" is the gravity, "theta" is projection angle.

☆Solution:

For Range to be maximum

$\implies \rm sin(2\theta) = 1$

We know that sin90=1

$\rm For\: sin(2\theta)\: to\:be\: 1 \\ \theta = 45\degree$

☆ Maximum range is achieved at angle 45°.

___________

At 45°

$\implies \rm Range = \dfrac{u^2 sin2(45\degree)}{g} = \dfrac{u^2 sin90}{g}$

$\longrightarrow \bf R = \dfrac{u^2}{g}$ ----(1)

$\implies \rm height = \dfrac{u^2 sin^245\degree}{2g} =\dfrac{u^2 (\dfrac{1}{\sqrt{2}})}{2g}$

$\longrightarrow \rm H =\dfrac{u^2}{2(2g)} = \dfrac{1}{4}\times \dfrac{u^2}{g}$

$\longrightarrow\bf H =\dfrac{1}{4}\times R$---{equation 1}

$\implies \bf R =4\times H$

Hence proved.

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