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3-2.Motion in Plane
medium
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
A
$\sqrt {H\,\cos \,\theta /g} $
B
$\sqrt {2H\,\cos \,\theta /g} $
C
$\sqrt {4H/g} $
D
$\sqrt {8H/g} $
(AIIMS-2013)
Solution
$\begin{array}{l}
Max.\,height = H = \frac{{{v^2}{{\sin }^2}\left( {90 – \theta } \right)}}{{2g}}\,\,\,….\left( i \right)\\
Time\,of\,flight,\,T = \frac{{2\,v\,\sin \left( {90 – \theta } \right)}}{g}\,\,\,\,\,…\left( {ii} \right)\\
From\left( i \right),\,\frac{{v\,\cos \,\theta }}{g} = \sqrt {\frac{{2H}}{g}} \,From\,\left( {ii} \right),\\
T = 2\sqrt {\frac{{2H}}{g}} = \sqrt {\frac{{8H}}{g}} .
\end{array}$
Standard 11
Physics
