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 :
\[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\]and\[T=\frac{2u\sin \theta }{g}\] \[\Rightarrow \] \[{{T}^{2}}=\frac{4{{u}^{2}}{{\sin }^{2}}\theta }{{{g}^{2}}}\] \[\therefore \] \[\frac{{{T}^{2}}}{H}=\frac{8}{g}\] \[\Rightarrow \]\[T=\sqrt{\frac{8H}{g}}=2\sqrt{\frac{2H}{g}}\]
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