A) \[\sqrt{\frac{2H}{g}}\]
B) \[2\sqrt{\frac{2H}{g}}\]
C) \[2\sqrt{\frac{2H\sin \theta }{g}}\]
D) \[\sqrt{\frac{2H\sin \theta }{g}}\]
Correct Answer: B
Solution :
Maximum height reached by the stone. \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2\,g}\] ?.. (i) where u is the velocity with which stone is thrown at angle 6. Time of flight, \[T=\frac{2u\,\sin \theta }{g}\] ... (ii) Using Eq. (i) \[{{u}^{2}}=\frac{2\,gH}{{{\sin }^{2}}\theta }\] \[\Rightarrow \] \[u=\frac{\sqrt{2\,gH}}{\sin \theta }\] Substituting the value of u in Eq. (ii), we obtain \[T=\frac{2}{8}\frac{\sqrt{2g\,H}}{\sin \theta }\times \sin \theta \] \[=\frac{8}{2}\times \sqrt{2\,g\,H}\] \[=2\sqrt{\frac{2H}{g}}\]
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