Aeroplane A and B are flying with constant velocity in the same vertical plane at angles 300 and 600 with respect to the horizontal respectively as shown in the figure. The speed of A is 1003 m/s. At time t=0 sec, an observer in A finds B at a distance of 500m. 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

For relative motion perpendicular to line of motion of AVA=1003=VBcos300VB=200m/st0=500VBsin300=500200×12=5sec

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Aeroplane A and B are flying with constant velocity in the same vertical plane at angles $30^{0}$ and $60^{0}$ with respect to the horizontal respectively as shown in the figure. The speed of A is $100 \sqrt{3} m / s$ . At time t=0 sec, an observer in A finds B at a distance of 500m. 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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For relative motion perpendicular to line of motion of A

$V_{A} = 100 \sqrt{3} = V_{B} \cos 30^{0}$

$V_{B} = 200 m / s$

$t_{0} = \frac{500}{V_{B} \sin 30^{0}} = \frac{500}{200 \times \frac{1}{2}} = 5 \sec$

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