Question

A particle $A$ 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 $B$ is projected in opposite direction with initial speed of $50\text{m/s}$ from a point at a distance of $100\text{m}$ from $A$. If the particles collide in air find the angle of projection $\alpha$ for particle $B$

• A. ${30}^{\circ }$
• B. ${37}^{\circ }$
• C. ${53}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is B ${37}^{\circ }$

Here, $a_A=-g\hat{j}$
$a_B=-g\hat{j}$
$u_{Ax}=60\cos {30}^{\circ }=30\sqrt{3}\text{m/s}$
$u_{Ay}=60\sin {30}^{\circ }=30\text{m/s}$
$u_{Bx}=-50\cos \alpha$
and $u_{By}=50\sin \alpha$
$\Rightarrow u_{AB}=(u_{Ax}-u_{Bx})\hat{i}+(u_{Ay}-u_{By})\hat{j}$
$\Rightarrow u_{AB}=(30\sqrt{3}+50\cos \alpha )\hat{i}+(30-50\sin \alpha )\hat{j}$
Relative acceleration between the two is zero as $a_A=a_B$. Hence, the relative motion between the two is uniform. It can be assumed that $B$ is at rest and $A$ is moving with $u_{AB}$. Hence, the two particles will collide, if $u_{AB}$ is along $AB$.

This is possible only when
$u_{Ay}=u_{By}$
i.e. component of relative velocity along $y-$ axis should be zero.
or $30=50\sin \alpha$
$\therefore \alpha ={\sin }^{-1}\left(\frac{3}{5}\right)={37}^{\circ }$