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 the figure- The speed of A is 100 -3 m - s- At time t -0 s- an observer in A finds B at a distance of 500 m- The 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 - t 0 in seconds is -

V_A = v_B cos30^∘ ⇒ 100√(3)=v_B×√(3)/2 ⇒ v_B=200 m/s ∴ t_0=d/v_B sin 30^∘ =500/200×1/2=5s. ...

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Byju's Answer

Standard XII

Physics

2D Motion as Two 1D Motions

Airplanes A a...

Question

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 the figure. The speed of A is $100 \sqrt{3}$ m/s. At time t = 0s, an observer in A finds B at a distance of 500m. The 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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Solution

$v_{A} = v_{B} c o s 30^{\circ}$

$\Rightarrow 100 \sqrt{3} = v_{B} \times \frac{\sqrt{3}}{2}$

$\Rightarrow v_{B} = 200 m / s$

$∴ t_{0} = \frac{d}{v_{B} s i n 30^{\circ}}$

$= \frac{500}{200 \times \frac{1}{2}} = 5 s$ .

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vA=vB cos30∘ ⇒100√3=vB×√32 ⇒vB=200 m/s ∴t0=dvB sin 30∘ =500200×12=5s.

$v_{A} = v_{B} c o s 30^{\circ}$

$\Rightarrow 100 \sqrt{3} = v_{B} \times \frac{\sqrt{3}}{2}$

$\Rightarrow v_{B} = 200 m / s$

$∴ t_{0} = \frac{d}{v_{B} s i n 30^{\circ}}$

$= \frac{500}{200 \times \frac{1}{2}} = 5 s$ .