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} $
D
$\sqrt {8H/g} $

✓

Answer

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

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