Question

A particle A is projected with an initial velocity of $60$ m/s at an angle ${30}^o$ to the horizontal. At the same time a second particle B is projected in opposite direction with initial speed of $50$m/s from a point at a distance of $100$m from A. If the particles collide in air, find the distance of P from A, where collision occurs. $(g=10m/s^2)$.

Answer 1

$uAx=60\cos {30}^o=30\sqrt{3}\frac{m}{s}$

$uAy=60\sin {30}^o=30\frac{m}{s}$

$uBx=-50\cos \alpha$

$uBy=50\sin \alpha$

A and B reach the same height at same time condition for collision

$uAy=uBy$

$30=50sin\alpha$

$\alpha ={sin}^{-1}\frac{3}{5}$