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# A partical A is projected with an initial velocity of $60 m/s$ at an angle $30 ^{\circ}$ to horizontal . At the same time a second particle B is projected in the opposite direction with initial speed of $50 m/s$ from a point at a distance of $100 m$ from A. If the particle collide in mid air. find the angle of projection $\alpha$ of particle B.$(g=10 m/s^2)$

$(a)\;\sin ^{-1} \frac{1}{2} \quad (b)\;\sin ^{-1} \frac{3}{4} \quad (c)\;\sin ^{-1} \frac{1}{3}\quad(d)\;\sin ^{-1} \frac{3}{5}$

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Taking X and Y direction as shown
$u_{Ax}=60 \cos 30 ^{\circ}$
$\qquad= 30 \sqrt 3 m/s$
$u_{Ay}=60 \sin 30=30 m/s$
$u_{Bx}=-50 \cos \alpha$
$u_{By}=50 \sin \alpha$
A and B reach the same height at same time condition for collision
$u_{Ay}=u_{By}$
$30=50 \sin \alpha$
$\alpha =\sin ^{-1} \bigg(\large\frac{3}{5}\bigg)$
Hence d is the correct answer.

edited Jan 26, 2014 by meena.p

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