Airplanes A and B are flying with constant velocity in the same vertical plane at angles 30° and 60° with respect to the horizontal respectively as shown in figure. The speed of A is 1003 ms−1. At time t=0 s, 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=t0, A just escapes being hit by B, t0 in seconds is

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Airplanes A and B are flying with constant velocity in the same vertical plane at angles $30 °$ and $60 °$ with respect to the horizontal respectively as shown in figure. The speed of A is $100 \sqrt{3} m s^{- 1} .$ At time $t = 0 s,$ 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

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answer is 5.

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Detailed Solution

Relative velocity of B with respect to A is perpendicular to line PA. Therefore, along line PA, velocity components of A and B should be same.
Question Image
$\Rightarrow v_{A} = 100 \sqrt{3} = v_{B} \cos 30 °$
$\Rightarrow v_{B} = 200 m s^{- 1}$
$t_{0} = \frac{500}{v_{B} \sin 30 °} = \frac{500}{200 \times 1 / 2} = 5 s$