Question
A projectile thrown with velocity $v$ making angle $\theta$ with vertical gains maximum height $H$ in the time for which the projectile remains in air, the time period is
✓
Answer
Max. height $= H = \frac{{{v^2}{{\sin }^2}\left( {90 - \theta } \right)}}{{2g}} .... (i)$
Time of flight $, T = \frac{{2\,v\,\sin \left( {90 - \theta } \right)}}{g} ...( {ii})$
From $\left( i \right),\,\frac{{v\,\cos \,\theta }}{g} $
$= \sqrt {\frac{{2H}}{g}} $ From left $(ii)$
$T = 2\sqrt {\frac{{2H}}{g}} $
$= \sqrt {\frac{{8H}}{g}} .$
Time of flight $, T = \frac{{2\,v\,\sin \left( {90 - \theta } \right)}}{g} ...( {ii})$
From $\left( i \right),\,\frac{{v\,\cos \,\theta }}{g} $
$= \sqrt {\frac{{2H}}{g}} $ From left $(ii)$
$T = 2\sqrt {\frac{{2H}}{g}} $
$= \sqrt {\frac{{8H}}{g}} .$
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