Eveningcat168
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A particles is projected at an angle of theta to a wall of height h metres standing on horizontaal ground at a distance c metres from the point of projection O. It just clears the wall at the highest point of its path. Prove that:
(a) the speed V of the projection is given by v^2 = g/2h(c^2 + 4h^2)
(b) the angle theta of projection is tan-1(2h/c)
many thanks
(a) the speed V of the projection is given by v^2 = g/2h(c^2 + 4h^2)
(b) the angle theta of projection is tan-1(2h/c)
many thanks
