A particle A is projected with an initial velocity of 60 m/s at an angle 30⁰ to the horizontal. At the same time a second particle B is projected in opposite direction with initial speed of 50 m/s from a point at a distance of 100 m from  A. If the particles collide in air, $\left(\text{g }=\text{ 1}0\text{m}/{\text{s}}^{\text{2}}\right)$ then

detailed solution

Correct option is B

Taking x and y directions as shown in figure.Hereacceleration of A=a→A=−gj^acceleration of B=a→B=−gj^horizontal component of initial velocity of A=uAx=60cos⁡30∘=303m/svertical component of initial velocity of A=uAy=60sin⁡30∘=30m/shorizontal component of initial velocity of B=uBx=−50cos⁡α vertical component of initial velocity of A=uBy=50sin⁡α   Relative acceleration between the two is zero . Hence, the relative motion between the two is uniform. Condition of collision is that it should be along AB. This is possible only when  component of relative velocity along y-axis is zero or  30 = 50 sinα ⇒  α= sin−13/5u→AB=uAx−uBx u→AB=(303+50cosα)m/s u→AB=303+50×45 u→AB=(303+40  )m/s  time of collision =t=ABu→AB t=100303+40   t=1.09s