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A particle has initial velocity =3i+4j and a constant force F=4i-3j acts on a particle the path of the particle is 1-straight line 2-parabolic 3-circular 4-elliptical A particle has initial velocity =3i+4j and a constant force F=4i-3j acts on a particle the path of the particle is 1-straight line2-parabolic3-circular4-elliptical
Here first we will find dot product. If a vector X = Ai+Bj and Y =Ci+Dj, dot product of X and Y = AC + BD so 3*4-4*3 = V*F*cosx So x is 90 This means velocity vector is perpendicular to accelaration and here accelaration is constant. This is the case of parabolic path when it is thrown horizontaly Hence it is parabolic.
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