A projectile thrown with velocity v making angle with vertical ga… - Vidyadip

MCQ

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} $
✓
$\sqrt {8H/g} $

✓

Answer

Correct option: D.

$\sqrt {8H/g} $

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

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