A ball is moving at 18ms\(^{-1}\) in a direction inclined at 60º to the horizontal. The horizontal component to its velocity is
Since the angle is inclined to the horizontal,
The horizontal component of the velocity V\(_x\) = V x cos \(\theta\) = 18 x cos 60º = 9m/s
(NOTE: Cos 60º = 0.5)
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