A particle A is projected with an initial velocity of $$60$$ $$m/s$$ at an angle $$300$$ 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, g $$=$$ $$10m/s2$$ then

Question

A particle A is projected with an initial velocity of $$60$$ $$m/s$$ at an angle $$300$$ 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, g $$=$$ $$10m/s2$$ then

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Taking x and y directions as shown in figure.Hereacceleration of $$A=a$$→$$A=$$−$$gj^acceleration$$ of $$B=a$$→$$B=$$−$$gj^horizontal$$ component of initial velocity of $$A=uAx=60cos$$⁡$$30$$∘$$=303m/svertical$$ component of initial velocity of $$A=uAy=60sin$$⁡$$30$$∘$$=30m/shorizontal$$ component of initial velocity of $$B=uBx=$$−$$50cos$$⁡α vertical component of initial velocity of $$A=uBy=50sin$$⁡α   Relative acceleration between the two is zero . Hence, the relative motion between the two is uniform. Condition of collision is that it should be along AB. This is possible only when  component of relative velocity along $$y-axis$$ is zero or  $$30$$ $$=$$ $$50$$ $$sin$$α ⇒  α$$=$$ $$sin$$−$$13/5u$$→$$AB=uAx$$−uBx u→$$AB=(303+50cos$$α$$)m/s$$ u→$$AB=303+50$$×$$45$$ u→$$AB=(303+40$$  $$)m/s$$  time of collision $$=t=ABu$$→AB $$t=100303+40$$   $$t=1.09s$$

A particle A is projected with an initial velocity of 60 m/s at an angle 300 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, g = 10m/s2 then

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