A) \[{{V}^{1}}\sin \theta '=V\,sin\theta \]
B) \[{{V}^{1}}\sin \theta '=-\,sin\theta \]
C) \[{{V}^{1}}\cos \theta '=V\,\cos \theta \]
D) \[{{V}^{1}}\cos \theta '=-V\,\cos \theta \]
Correct Answer: A
Solution :
[a] As the floor exerts a force on the ball along the normal, & no force parallel to the surface, therefore the velocity component along the parallel to the floor remains constant. \[\operatorname{Hence} V sin \theta ={{V}^{1}} sin {{\theta }^{1}}.\]
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