A) \[\sqrt{\frac{2H}{g}}\]
B) \[2\sqrt{\frac{2H}{g}}\]
C) \[\frac{2\sqrt{2H\sin \theta }}{g}\]
D) \[\frac{\sqrt{2H\sin \theta }}{g}\]
Correct Answer: B
Solution :
Maximum height reached by the stone. \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] ?(i) where \[u\] is the velocity with which stone is thrown at angle\[\theta .\] Time of flight, \[T=\frac{2u\sin \theta }{g}\] ?(ii) Using Eq. (i) \[{{u}^{2}}=\frac{2gH}{{{\sin }^{2}}\theta }\] \[\Rightarrow \] \[u=\frac{\sqrt{2gH}}{\sin \theta }\] Substituting the value of \[u\]in Eq. (ii), we obtain \[T=\frac{2}{g}\frac{\sqrt{2gH}}{\sin \theta }\times \sin \theta \] \[=\frac{2}{g}\times \sqrt{2gH}\] \[=2\sqrt{\frac{2H}{g}}\]
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