Two particles are projected simultaneously from the top of a tower. The first particle is projected horizontally at speed $$u1$$ $$=$$ $$10$$ $$m/s$$ and the other one is projected at an angle θ$$=$$$$tan$$−$$1$$⁡$$43$$, to the horizontal with a velocity $$u2$$. Find minimum value of $$u2(in$$ $$m/s)$$ so that the velocity vector of the two particles can get perpendicular to each other at some point of time during their course of flight (g$$ $$=$$ $$10$$ $$m/s2).

Question

Two particles are projected simultaneously from the top of a tower. The first particle is projected horizontally at speed $$u1$$ $$=$$ $$10$$ $$m/s$$ and the other one is projected at an angle θ$$=$$$$tan$$−$$1$$⁡$$43$$, to the horizontal with a velocity $$u2$$. Find minimum value of $$u2(in$$ $$m/s)$$ so that the velocity vector of the two particles can get perpendicular to each other at some point of time during their course of flight (g$$ $$=$$ $$10$$ $$m/s2).

Infinity Learn · Accepted Answer

The velocity of the particles after time 't', using v→$$=u$$→$$+a$$→tThe velocity of the $$1st$$ particle, $$v1$$→$$=10i^$$−$$gtj^=10i^$$−$$10tj^The$$ velocity of the $$2nd$$ particle, $$v2$$→$$=u2cos$$⁡θ$$i^+u2sin$$⁡θ−$$gtj^=35u2i^+45u2$$−$$10tj^For$$ the two velocities to be perpendicular, $$v1$$→⋅$$v2$$→$$=0$$∴$$10$$⋅$$35u2$$−$$10t45u2$$−$$10t=0$$⇒$$3u2$$−$$t4u2$$−$$50t=0$$⇒$$50t2$$−$$4u2t+3u2=0For$$ this quadratic equation to have real solutions, D≥$$0$$⇒$$4u22$$−$$4$$×$$50$$×$$3u2$$≥$$0$$⇒$$u2$$≥$$752m/s$$    ∴$$umin=752=37.5m/s$$

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