The correct option is
B 2√3
Step 1: Condition of collision
Clearly from figure, the particles can collide at the maximum height of the projectile.
Let the particles collide in time t.
Since the projectile motion is symmetric, So time to reach maximum height t=T2, Where T is the time of flight of the projectile.
Till collision, distance travelled by B equals the Maximum height of the projectile, i.e. h
Step 2 : Maximum height and Time Calculation
Taking upward as positive
Maximum height travel by the particle A is h=u2 sin2θg=V2Asin260o2g =3V2A8g ....(1)
Step 3 : Distance traveled by the second particle in time t
For Particle A:
Time of flight T=2u sinθg=2VAsin60og
Now, t=T/2=VAsin60og=VA√32g
For Particle B:
Since acceleration is constant, therefore applying equation of motion in y direction, (Taking upward as positive)
S=VBt−12gt2 =VB(VA√32g)−12g×V2A×34g2 ....(2)
Step 3: Calculations
Since distance traveled by B equal the maximum height of projectile, So using Equations (1) and (2)
∴S=h
⇒VB(VA√32g)−12g×V2A×34g2=3V2A8g⇒VAVB=2√3
Hence, Option B is correct.