Two balls are thrown horizontally from the top of a building with speed u₁ and u₂ respctively in opposite deirections. The seperation between two balls when they are moving perpendicular to each other, is ...........

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Please find below the solution to the asked query:

$$\begin{gathered} \text{Suppose the balls will be perpendicular after a time }t.\text{ So after time t the} \\ \text{velocity vector of the balls will be respectively } \\ {\overrightarrow{v}}_1=u_1\hat{i}-gt\hat{j} \\ \text{and } \\ {\overrightarrow{v}}_2=-u_2\hat{i}-gt\hat{j} \\ \text{Now if they are moving at perpendicular direction, then angle between their } \\ \text{velocities will be} 90°. H\text{ence the dot product of this two vector will be zero. } \\ \text{Hence we have} \\ {\overrightarrow{v}}_1.{\overrightarrow{v}}_2=0=\left(u_1\hat{i}-gt\hat{j}\right).\left(-u_2\hat{i}-gt\hat{j}\right) \\ \Rightarrow -u_1{u}_2+g^2{t}^2=0 \\ \Rightarrow t=\pm \frac{\sqrt{u_1{u}_2}}{g} \\ \text{We will take only positive time as negative time has no meaning. } \\ \text{Now since the initial vertical velocity and vertical acceleration are same for } \\ \text{both the ball, so the seperation will come from the horizontal velocity. So} \\ \text{the separation will be } \\ s=(u_1+u_2)t \text{(since the balls are moving in opposite direction)} \\ =(u_1+u_2)\frac{\sqrt{u_1{u}_2}}{g} \end{gathered}$$

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