1HINT:
v=\omegar
the trajectory followed is that of parabola
v vertical(y axis) is 0 at the highest point
v horizontal remains unchanged
therefore \omega=v(horizontal)/H
so basically u have to find u horizontal...
A particle is projected from ground with speed "u" such that the horizontal range is "R" and the maximum height is "H". The angular velocity of the particle about the point of projection, when it is at the highest point of its trajectory is:
(A) HRu/(R2 + 4H 2)√R2 +16H2
(B) 4HRu/(R2 + 4H2 )√R2 +16H2
(C) u/H
(D) uR/H √R2 +16H2
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3 Answers
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