Three particles are projected in the air with the minimum possible...

For a given speed,

the maximum range of a projectile on a horizontal level is

{R}_\max = \dfrac{u2}{g}

the maximum range up and down the incline are

\left({R}_{up}\right)max = \dfrac{u2}{g\left(1 + sin\alpha\right)}\left({R}_{down}\right)max = \dfrac{u2}{g\left(1 - sin\alpha\right)}$</p><p>by substituting the values of the velocities given in the problem, we get</p><p>$R = {u}_{2}^{2}g\left(1 + sin \alpha\right) ⇒ {u}_{2}^{2} = Rg\left(1 + sin \alpha\right) \ldots ( 1 )$<br />$2Rcos \alpha = {u}_{3}^{2}g ⇒ {u}_{3}^{2} = 2Rgcos \alpha \ldots ( 2 )$<br />Now,${u}_{1}{u}_{2} = Rg\sqrt{1 - \sin^{2} \alpha} = Rgcos\alpha = \dfrac{{u}_{3}^{2}}{2}$<br />$∴ {u}_{3}^{2} = 2{u}_{1}{u}_{2}$