MCQ
Airplanes $A$ and $B$ are flying with constant velocity in the same vertical plane at angles $30^{\circ}$ and $60^{\circ}$ with respect to the horizontal respectively as shown in figure . The speed of $A$ is $100 \sqrt{3} ms ^{-1}$. At time $t=0$, an observer in A finds $B$ at a distance of $500 \ m$. This observer sees $B$ moving with a constant velocity perpendicular to the line of motion of $A$. If at $t = t _0$, $A$ just escapes being hit by $B , t _0$ in seconds is:
  • $5$
  • B
    $6$
  • C
    $7$
  • D
    $8$

Answer

Correct option: A.
$5$
a
For relative motion perpendicular to line of motion of $A$ $A$

$V_A=100 \sqrt{3}=V_B \operatorname{Cos} 30^{\circ} $

$\Rightarrow \quad V_B=100 m / s $

$t_0=\frac{50}{V_B \sin 30^{\circ}}=\frac{500}{200 \times \frac{1}{2}}=5 sec \text { Ans. }$

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