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

Taking x and y directions as shown in figure.

Here
$\begin{array}{l}acceleration of A={\overrightarrow{a}}_A=-g\widehat{j}\\ acceleration of B={\overrightarrow{a}}_B=-g\widehat{j}\\ horizontal component of initial velocity of A=u_{Ax}=60\cos {30}^{\circ }=30\sqrt{3}m/s\\ vertical component of initial velocity of A=u_{Ay}=60\sin {30}^{\circ }=30m/s\\ horizontal component of initial velocity of B=u_{Bx}=-50\cos \alpha \\ vertical component of initial velocity of A=u_{By}=50\sin \alpha \end{array}$
   
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

  $$\begin{gathered} \text{3}0=\text{ 5}0\text{ sin}\alpha \\ \Rightarrow \alpha ={\text{ sin}}^{-\text{1}}\left(\text{3}/\text{5}\right) \end{gathered}$$

$$\begin{gathered} \left|{\overrightarrow{u}}_{AB}\right|=u_{Ax}-u_{Bx} \\ \left|{\overrightarrow{u}}_{AB}\right|=(30\sqrt{3}+50\cos \alpha )m/s \\ \left|{\overrightarrow{u}}_{AB}\right|=\left(30\sqrt{3}+50\times \frac{4}{5}\right) \\ \left|{\overrightarrow{u}}_{AB}\right|=(30\sqrt{3}+40 )m/s \end{gathered}$$

 $$\begin{gathered} time of collision =t=\frac{AB}{\left|{\overrightarrow{u}}_{AB}\right|} \\ t=\frac{100}{30\sqrt{3}+40} \\ t=1.09s \end{gathered}$$