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Hi,
Concept: V(cross)A is a vector perpendicular to the plane containing V(vector) and (A vector).
Consider horizontal plane as the X-Y plane, so that gravity is perpendicular to this plane (along -ve Z-direction)
And this vector froce, F should cancel out the vecotr force due to gravity.
Let's assume that gravity is in the (Z - axis direction), which would be = mg(downwards)
Now the force you apply should exactly cancel out this force. Hence F(vector) should be in the upwards direction with magnitude = mg.
Mag |F| = mg = VAsin(Q); where Q is the angle between the vector V and vector A. (Both V and A in the horizontal plane X-Y), so that their cross product F is in the vertical direction.
So |V| should be mg/(|A|*sinQ)
now for the minimum speed, sinQ should be maximum (which we know is 1), hence Q = 90 degrees.
Hence V(vector) should be perpendicular to A(vector) in the horizontal plane.
And hence if A is in the posive x-direction, V should be along the negative Y-direction, so that V(cross)A is along the positive Z-direction, to cancel out the effect of gravity along the negative Z-direction.
Hope that was clear and helpful.
All the best
Regards,
Ashwin (IIT Madras)
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