A projectile thrown with velocity v making an | vedclass

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Question

$\begin{array}{l}
Max.\,height = H = \frac{{{v^2}{{\sin }^2}\left( {90 - 	heta } ight)}}{{2g}}\,\,\,....\left( i ight)\
Time\,of\,flight,\,T =

Vedclass · Accepted Answer

$\sqrt {8H/g} $

Vedclass · Answer

$\begin{array}{l}
Max.\,height = H = \frac{{{v^2}{{\sin }^2}\left( {90 - 	heta } ight)}}{{2g}}\,\,\,....\left( i ight)\
Time\,of\,flight,\,T = \frac{{2\,v\,\sin \left( {90 - 	heta } ight)}}{g}\,\,\,\,\,...\left( {ii} ight)\
From\left( i ight),\,\frac{{v\,\cos \,	heta }}{g} = \sqrt {\frac{{2H}}{g}} \,From\,\left( {ii} ight),\
T = 2\sqrt {\frac{{2H}}{g}}  = \sqrt {\frac{{8H}}{g}} .
\end{array}$

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

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