Question

Three particles are projected in the air with the minimum possible speeds, such that the first goes from $A$ to $B$ , the second goes from $B$ to $C$ and the third goes from $C$ to $A$ . Points $A$ and $C$ are at the same horizontal level. The two inclines make the same angle $\alpha $ with the horizontal, as shown. The relation among the projection speeds of the three particles is

• A. $u_{3}=u_{1}+u_{2}$
• B. $u_{3}^{2}=2u_{1}u_{2}$
• C. $\frac{1}{u_{3}}=\frac{1}{u_{1}}+\frac{1}{u_{2}}$
• D. $u_{3}^{2}=u_{1}^{2}+u_{2}^{2}$

Answer 1

Answer: (B)

For a given speed,
the maximum range of a projectile on a horizontal level is
$R_{\max}=\frac{u^{2}}{g}$
the maximum range up and down the incline are
$\left(R_{\text{up}}\right)_{\max}=\frac{u^{2}}{g \left(1 + \sin \alpha \right)}\left(R_{\text{down}}\right)_{\max}=\frac{u^{2}}{g \left(1 - \sin \alpha \right)}$
by substituting the values of the velocities given in the problem, we get
$R=\frac{u_{2}^{2}}{g \left(1 + \sin \alpha \right)}\Rightarrow u_{2}^{2}=Rg\left(1 + \sin ⁡ \alpha \right)\ldots \left(\right.1\left.\right)$
$2R\cos \alpha = \, \frac{u_{3}^{2}}{g}\Rightarrow u_{3}^{2} \, =2Rg\cos⁡\alpha \ldots \left(\right.2\left.\right)$
Now, $u_{1}u_{2}=Rg\sqrt{1 - \sin^{2} \alpha }=Rg \, \cos \, \alpha =\frac{u_{3}^{2}}{2}$
$\therefore u_{3}^{2} \, =2u_{1}u_{2}$