A particle $\text{A}$ is projected with an initial velocity of $60\text{m/s}$ at an angle ${30}^{\circ }$ to the horizontal. At the same time, a second particle $\text{B}$ is projected in the direction as shown in the figure with an initial speed of $50\text{m/s}$ from a point at a distance of $100\text{m}$ from $\text{A}$. If the particles collide in the air, find the time when the particles collide.

The correct option is A $1.09\text{s}$

As both the particles collide in the air, so

Vertical height covered by the particle $\text{A}$ in time $t=$ Vertical height covered by the particle $\text{B}$ in time $t$

$\Rightarrow h_A=h_B$

$\Rightarrow u_{Ay}t-\frac{1}{2}g{t}^2=u_{By}t-\frac{1}{2}g{t}^2$

$\Rightarrow u_{Ay}=u_{By}$

$\Rightarrow u_A\sin {30}^{\circ }=u_B\sin \alpha$

$\Rightarrow \sin \alpha =\frac{3}{5}$ or $\cos \alpha =\frac{4}{5}$


Taking $x$ and $y$ directions as shown in figure,

$|u_{AB}|=u_{Ax}-u_{Bx}=60\cos {30}^{\circ }-(-50\cos \alpha )$
$|u_{AB}|=(30\sqrt{3}+50\times \frac{4}{5})\text{m/s}$
$|u_{AB}|=(30\sqrt{3}+40)\text{m/s}$

Therefore, the time of collision is ,

$t=\frac{AB}{|u_{AB}|}=\frac{100}{30\sqrt{3}+40}=1.09\text{s}$

Hence, option $\left(\text{A}\right)$ is correct.