A) \[\sqrt{\frac{(2h\,\sin \,\theta }{g}}\]
B) \[2\sqrt{\frac{(2h\,\sin \,\theta }{g}}\]
C) \[2\sqrt{\frac{(2h)}{g}}\]
D) \[\sqrt{\frac{(2h)}{g}}\]
Correct Answer: C
Solution :
Maximum height attained by projectile is \[h=\frac{{{u}^{2}}\sin \theta }{2g}\] ?(i) \[\therefore \] \[u\sin \theta =\sqrt{2gh}\] ?(ii) Time of flight is given by \[T=\frac{2u\sin \theta }{g}\] ?(iii) From Eqs. (ii) and (iii), we get the time of flight is \[=\frac{2\sqrt{2gh}}{g}=2\sqrt{\frac{2gh}{{{g}^{2}}}}=2\sqrt{\frac{2h}{g}}\]
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