Question

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

• A. $1.09s$
• B. $1.38s$
• C. $1.53s$
• D. $1.72s$

Answer 1

The correct option is A $1.09s$


As both the particles collide in the air, so
Vertical height covered by the particle $A$ in time $t=$ Vertical height covered by the particle $B$ 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
$|u_{AB}|=u_{Ax}-u_{Bx}=60\cos {30}^{\circ }-(-50\cos \alpha )$
$=(30\sqrt{3}+50\times \frac{4}{5})\text{m/s}$
$=(30\sqrt{3}+40)m/s$
Therefore, time of collision is $t=\frac{AB}{|u_{AB}|}=\frac{100}{30\sqrt{3}+40}$ or $t=1.09s$