A particle is projected on a frictionless inclined plane of...

While going upward the time of flight is

$T = \dfrac{2usin \alpha}{gcos \beta}$

So the range on the inclined plane is

$R = \left(ucos \alpha\right)T - \dfrac{1}{2}\left(gsin \beta\right){T}^{2}$

Just after the collision of the projectile with the inclined plane, the components of its velocity down the inclined plane and perpendicular to the inclined plane are

${v}_{∥} = \left(gsin \beta\right)T - ucos \alpha$, ${v}_{⊥} = eusin \alpha$

The time of flight while going down the inclined plane is

${T}^{'} = \dfrac{2eusin \alpha}{gcos \beta} = eT$

The range of the projectile while going down is

$R = \left[\left(gsin \beta\right)T - ucos \alpha\right]eT + \dfrac{1}{2}\left(gsin \beta\right){e}^{2}{T}^{2}$

$⇒ \left[\left(gsin \beta\right)T - ucos \alpha\right]eT + \dfrac{1}{2}\left(gsin \beta\right){e}^{2}{T}^{2} = \left(ucos \alpha\right)T - \dfrac{1}{2}\left(gsin \beta\right){T}^{2}$

$⇒ ucos \alpha\left(1 + e\right) = \left(gsin \beta\right)T\left[\dfrac{1}{2} + e + \dfrac{{e}^{2}}{2}\right]$

Substituting the value as $T = \dfrac{2usin \alpha}{gcos \beta}$ we get

$⇒ cot \alpha\left(1 + e\right) = 2tan \beta\left[\dfrac{1}{2} + e + \dfrac{{e}^{2}}{2}\right]$

$⇒ 3{e}^{2} + 2e - 1 = 0$

$⇒ e = \dfrac{1}{3}$ or $\dfrac{1}{e} = 3$