Physics - Kinematics Question with Solution | TestHub

While going upward the time of flight is

$\mathrm{T}=\frac{2\mathrm{usin\alpha }}{\mathrm{gcos\beta }}$

So the range on the inclined plane is

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

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

${\mathrm{v}}_{\parallel }=\left(\mathrm{gsin\beta }\right)\mathrm{T}-\mathrm{ucos\alpha }$, ${\mathrm{v}}_{\perp }=\mathrm{eusin\alpha }$

The time of flight while going down the inclined plane is

${\mathrm{T}}^{\text{'}}=\frac{2\mathrm{eusin\alpha }}{\mathrm{gcos\beta }}=\mathrm{eT}$

The range of the projectile while going down is

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

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

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

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

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

$\Rightarrow 3{\mathrm{e}}^2+2\mathrm{e}-1=0$

$\Rightarrow \mathrm{e}=\frac{1}{3}$ or $\frac{1}{\mathrm{e}}=3$