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
$R=\frac{u_2^2}{g\left(1+\sin \alpha \right)}\Rightarrow u_2^2=Rg\left(1+\sin \alpha \right)\dots (1)$
$2R\cos \alpha =\frac{u_3^2}{g}\Rightarrow u_3^2=2Rg\cos \alpha \dots (2)$
Now, $u_1{u}_2=Rg\sqrt{1-{\sin }^2\alpha }=Rg\cos \alpha =\frac{u_3^2}{2}$
$\therefore u_3^2=2u_1{u}_2$