A) \[T=\frac{v_{0}^{2}\sin 2\theta }{g}\]
B) \[T=\frac{v_{0}^{2}\,\,{{\sin }^{2}}\theta }{2g}\]
C) \[T=\frac{v_{0}^{2}}{g}\]
D) \[T=\frac{2{{v}_{0}}\,\sin \,\theta }{g}\]
Correct Answer: D
Solution :
Time of flight, \[T=\frac{2{{v}_{0}}\,\sin \theta }{g}\]
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