A stone is thrown at an angle $\theta $ to the horizontal reaches a maximum height $H$. Then the time of flight of stone will be
A stone is thrown at an angle $\theta $ to the horizontal reaches a maximum height $H$. Then the time of flight of stone will be
$\sqrt {\frac{{2H}}{g}} $
$2\,\sqrt {\frac{{2H}}{g}} $
$\frac{{2\sqrt {2H\,\sin \theta } }}{g}$
$\frac{{\sqrt {2H\,\sin \theta } }}{g}$
A stone is thrown at an angle $\theta $ to the horizontal reaches a maximum height $H$. Then the time of flight of stone will be
$H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}$ and $T = \frac{{2u\sin \theta }}{g}$
$⇒$ ${T^2} = \frac{{4{u^2}{{\sin }^2}\theta }}{{{g^2}}}$
$\frac{{{T^2}}}{H} = \frac{8}{g}$
$⇒$ $T = \sqrt {\frac{{8H}}{g}} = 2\sqrt {\frac{{2H}}{g}} $