Question

Airplanes A and B are flying with constant velocity in the same vertical plane at angles 30 and 60° withrespect to the horizontal respectively as shown in figure. The speed of A is 100 /3 ms. At time t = Os,an observer in A finds B at a distance of 500m. This observer sees B moving with a constant velocityperpendicular to the line of motion of A. If at t = to, Ajust escapes being hit by B, to in seconds is:(JEE (Advanced) 2014, P-1,3/60]​

Answer 1

Answer:

The answer will be 25.98 s

Explanation:

According to the problem two planes have a constant velocity.

Plane A is making angle of 30° and B is making angle of 60°.

The speed of A is 100/3 ms and the distance between A and B is 500 m

Now, the velocity of plane B V(B) has two components V(B)cos30° and V(B)sin30° as it is making angle of 60° with the plane.

Now according to the problem we can say that, V(A) = V(B)cos30° 100/3 = V(B) √3/2 => V(B) = 38.49 m/s

Now we need to find t0

we know time = distance / velocity

t0 = 500/V(B)sin30° as B is moving perpendicular to the line of A

= 500/38.49 x 1/2 = 25.98 s

Answer 2

V(b)cos 30 = V(a)

V(b) = 100√3 ÷ cos30

V(b) = 200m/s

t0 = Relative distance/Relative velocity

t0 = 500/V(B)sin30

= 5 second