A) \[\sqrt{\frac{2\,\,H}{g}}\]
B) \[2\sqrt{\frac{2\,\,H}{g}}\]
C) \[\frac{2\sqrt{2H\sin \theta }}{g}\]
D) \[\frac{\sqrt{2H\sin \theta }}{g}\]
Correct Answer: B
Solution :
Maximum height \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] ... (1) Time of flight \[T=\frac{2u\sin \theta }{g}\] \[\Rightarrow \] \[u\sin \theta =\frac{Tg}{2}\] ... (2) From equations (1) and (2), we get \[H=\frac{1}{2g}{{\left( \frac{Tg}{2} \right)}^{2}}\] \[H=\frac{{{T}^{2}}{{g}^{2}}}{8g}\] \[{{T}^{2}}=\frac{8H}{g}\] \[T=2\sqrt{\left( \frac{2H}{g} \right)}\]
You need to login to perform this action.
You will be redirected in
3 sec
