A particle is projected at an angle of 60^(@) with the horizontal from the ground with a velocity 10sqrt3ms^(-1). The angle between velocity vector after 2s and initial velocity vector is (g=10ms^(_2))

A particle is projected at an angle of 60^(@) with the horizontal from

1.	A particle is projected at an angle of 60^(@) with the horizontal from the ground with a velocity 10sqrt3ms^(-1). The angle between velocity vector after 2s and initial velocity vector is (g=10ms^(_2))
Answer» `0^(@)` `30^(@)` `60^(@)` `90^(@)` Solution :Initial VELOCITY, `v_(i)=2costhetahati+4sinthetahatj=5sqrt3hati+15hatj` Final velocity VECTOR (after 2S), `v_(f)=ucosthetahati+(usintheta-"GT")hatj=5sqrt3hati+5hatj` Now, `v_(i)*v_(f)=25xx3-15xx5=0` `thereforev_(i)botv_(f)`

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A particle is projected at an angle of 60^(@) with the horizontal from the ground with a velocity 10sqrt3ms^(-1). The angle between velocity vector after 2s and initial velocity vector is (g=10ms^(_2))

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