WBJEE 2019 Physics Solved Paper

According to the question,In two dimension, the position vectors r1 and r2 represented as $r_{1}=v t \hat{i} - \frac{1}{2} g t^{2} \hat{j} \dots (i)$ $r_{2}=v t(-\hat{i}) - \frac{1}{2} g t^{2}(\hat{j}) \dots (ii)$ ∵ We know that, when the two vectors are mutually perpendicular, i.e. $\theta = 90^{\circ}$ So $r_{1} \cdot r_{2} = r_{1} r_{2} \cos 90^{\circ}$ $r_{1} \cdot r_{2} = 0$ Substituting the values $r_{1}$ and $r_{2}$ in the above relation, we get $-v^{2} t^{2} + \frac{1}{4} 4 g^{2} t^{4} = 0$ (where, $.\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1)$ $v^{2} t^{2} = \frac{1}{4} g^{2} t^{4}$ $\Rightarrow v^{2} = \frac{1}{4} g^{2} t^{2}$ ∴ Magnitude of velocity of the particles, $v = \frac{1}{2} g t$ We know that, separation distance between particles at a time $t$ $\Delta x = 2 v t$ $\Delta x = 2 \times v \times \frac{2v}{g} \Rightarrow \Delta x = \frac{4 y^{2}}{g}$