Question

Two particles are simultaneously thrown from the top of two towers as shown. Their velocities are $2m{s}^{-1}$ and $14m{s}^{-1}$ . Horizontal and vertical separations between these particles are $22m$ and $9m$ respectively. Then the minimum separation between the particles in the process of their motion in meters is ( $g=10m{s}^{-2}$ )

Answer 1

Answer: 6

$v_x=8\sqrt{2}m{s}^{-1}=$ relative velocity along $x$-axis
$\therefore x=22-8\sqrt{2}t$
$v_y=6\sqrt{2}m/s=$ relative velocity
along the $y$-axis
$\therefore y=9-6\sqrt{2}t$
$r=\sqrt{x^2+y^2}$
For minimum $r$,
$\frac{dr}{dt}=0\Rightarrow t=\frac{23}{10\sqrt{2}}s$
$r=\sqrt{x^2+y^2}$
substituting the value of $t$ in $r,r_{\min }$ $=6m$