Two particles are simultaneously thrown from the roofs of two high buildings as shown in figure. Their velocities are  ${\text{V}}_{\text{A}}\text{= 2 m}/{\text{s\hspace{0.17em}\hspace{0.17em}and V}}_{\text{B}}=\text{14 m}/\text{s}$ respectively. Then

Assuming A to at rest  ${\overline{a}}_{BA}={\overline{a}}_B-{\overline{a}}_A=0$ as ${\overline{a}}_A={\overline{a}}_B-g$  (downwards)
Thus, the relative motion between them is uniform. Relative velocity of B with respect to A in vertical direction, and horizontal direction
Relative velocity of B with respect to A in horizontal direction.
${\text{u}}_{\text{BAV}}={\text{ u}}_{\text{B}}{\text{sin 45-u}}_{\text{A}}\text{sin 45 }=\left(\text{12}\right)\frac{1}{\sqrt{2}}=6\sqrt{2}m/s$
Relative velocity of B with respect to A in vertical direction.
$$\begin{gathered} {\text{u}}_{\text{BAH}}={\text{ u}}_{\text{B}}{\text{cos 45}}^0{\text{ -\hspace{0.17em}(-u}}_{\text{A}}{\text{cos 45}}^0\text{)}=\left(\text{14+2}\right)\frac{1}{\sqrt{2}}=8\sqrt{2}m/s \\ \tan \theta =\frac{U_{ABV}}{U_{ABH}}=\frac{6\sqrt{2}}{8\sqrt{2}}=\frac{3}{4} \\ \theta =37° \end{gathered}$$
Horizontal distance between A and B after time t is
$x=22-\left(u_{BAH}\right)t=(22-8\sqrt{2}t)m$
and vertical distance between A and B time t is
$y=9-u_{BAV}t=\left(9-6\sqrt{2}t\right)m$
Therefore, distance between them after time t is  $s=\sqrt{x^2+y^2}$
For S to be minimum  $\frac{d}{dt}\left(s^2\right)=0$
$\therefore (22-8\sqrt{2t})(-8\sqrt{2})+2(9-6\sqrt{2}t)(-6\sqrt{2})=0$
Therefore   $t=\frac{23}{10\sqrt{2}S}$

$$\begin{gathered} \tan 37=\frac{x+9}{22} \\ \frac{3}{4}=\frac{x+9}{22} \\ x=7.5m \\ \sin 53=\frac{x\text{'}}{x} \\ \frac{4}{5}=\frac{x\text{'}}{7.5} \\ x\text{'}=6m \end{gathered}$$

Paragraph- I
A particle is projected horizontally with a speed V = 5 m/s from the top of a plane inclined at an angle $\theta ={37}^0$  to the horizontal. (g = 10 m/$s^2$)