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 at an angle $\alpha$ to the horizontal with an initial speed of $50m/s$ from a point at a distance of $100m$ from $A$ as shown in the figure. If the particles collide in the air, find the angle of projection $\alpha$ of particle $B$.

• A. ${30}^{\circ }$
• B. ${60}^{\circ }$
• C. $\sin {}^{-1}\left(\frac{3}{5}\right)$
• D. $\cos {}^{-1}\left(\frac{3}{5}\right)$

Answer 1

The correct option is C $\sin {}^{-1}\left(\frac{3}{5}\right)$


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

Here, $\overrightarrow{a_A}=-g\hat{j}$ and $\overrightarrow{a_B}=-g\hat{j}$
$u_{Ax}=60\cos {30}^{\circ }=30\sqrt{3}m/s$
$u_{Ay}=60\sin {30}^{\circ }=30m/s$
$u_{Bx}=-50\cos \alpha$ and
$u_{By}=50\sin \alpha$
Relative acceleration between the two is zero as $\overrightarrow{a_A}=\overrightarrow{a_B}$. Hence, the relative motion between the two is uniform.
It can be assumed that $B$ is at rest and $A$ is moving with ${\overrightarrow{u}}_{AB}$. Hence, the two particles will collide if ${\overrightarrow{u}}_{AB}$ is along line $AB$.
This is posible only when,
$u_{Ay}=u_{By}$
(i.e. component of relative velocity along $y-$axis must be zero)
$⟹60\sin {30}^{\circ }=50\sin \alpha$
$⟹\alpha ={\sin }^{-1}\left(\frac{3}{5}\right)$