Airplanes $A$ and $B$ are in the same vertical plane and flying with constant velocity. At time $t=0\text{sec}$, an observer in $A$ finds $B$ at a distance of $500\text{m}$ and moving perpendicular to the line of motion of $A$. If at $t=t_0\text{sec}$, $A$ just escapes being hit by $B$ , $t_0$ in seconds is

Let the velocity of airplane $B$ be $v$.


As observer in $A$ finds $B$ moving perpendicular to the line of motion of $A$ i.e perpendicular to $OX$.
Taking the $x\&y$ axis as shown in figure.
Velocity of $B$ w.r.t $A$ along the line of motion
$(v_{BA}{)}_{OX}=0$
$\Rightarrow (v_B{)}_{OX}-(v_A{)}_{OX}=0$
$\Rightarrow v\cos {30}^{\circ }-100\sqrt{3}=0$
$\Rightarrow v=200\text{m/s}...\left(1\right)$
Also, distance to be travelled by $B$ along the perpendicular direction to line of motion of $A$ to hit $A$ is,
$d=500\text{m}$.
$\therefore$ Time taken $=\frac{d}{v_{OY}}$
where $v_{OY}$ is the component of velocity of $B$ in a direction perpendicular to the motion of $A$,
$t=\frac{d}{v\sin {30}^{\circ }}$
$\Rightarrow t=\frac{500}{200\sin {30}^{\circ }}$
$\therefore t=5\text{sec}$
Hence time at which $A$ just escapes colliding with $B$ is $t_0=5\text{sec}$