Question

A projectile has the same range R when the maximum height attained by it is either H1 or H2. Find the relation between R, H1 and H2​

Answer 1

$\underline{\pink{\mathfrak{Answer:-}}}$

$\boxed{\mathsf{R = 4\sqrt{H1H2}} }$

$\underline{\pink{\mathfrak{Answer:-}}}$

Given

Range is same for a projectile, when it attains the maximum height H1 and H2

It means, range is same for Φ and (90-Φ) angles

To Find

relation between R ,H1 and H2

Solution

$\mathsf{H1 = \dfrac{{u}^{2}{sin}^{2} \theta}{2g}}$ ------(1)

$\mathsf{H2 = \dfrac{{u}^{2}{sin}^{2} (90-\theta)}{2g}}$

$\mathsf{H2 = \dfrac{{u}^{2}{cos}^{2} \theta}{2g}}$--------(2)

From [1] and [2]

$\mathsf{H1H2 = \dfrac{{u}^{2}{sin}^{2} \theta}{2g} \times \dfrac{ {u}^{2}{cos}^{2} \theta}{2g}}$

$\mathsf{H1H2 = \dfrac{{u}^{4}{cos}^{2} \theta {sin}^{2}\theta}{4{g}^{2}}}$

$\mathsf{H1H2 = {[\dfrac{{u}^{2}cos \theta sin\theta}{2g}]}^{2}}$

$\mathsf{\sqrt{H1H2} = \dfrac{{u}^{2}cos \theta sin \theta}{2g}}$

$\mathsf{\sqrt{H1H2} = \dfrac{{u}^{2}2 cos \theta sin \theta}{4g}}$

$\mathsf{\sqrt{H1H2} = \dfrac{{u}^{2}sin 2\theta}{4g}}$

$\mathsf{\sqrt{H1H2} = \dfrac{{u}^{2}sin 2\theta}{g} \times \dfrac{1}{4}}$

$\mathsf{\sqrt{H1H2} = R \times \dfrac{1}{4}}$

$\boxed{\red{\mathsf{R = 4\sqrt{H1H2} }}}$