Question
A projectile is thrown with velocity $v$ at an angle $\theta$ with horizontal. When the projectile is at a height equal to half of the maximum height, the vertical component of the velocity of projectile is ...........
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Answer
(c)
$v_B^2=v^2 \sin ^2 \theta-\frac{2 g}{2}\left(\frac{u^2 \sin ^2 \theta}{2 g}\right)$
$v_B^2=\frac{v^2 \sin ^2 \theta}{2}$
$v_B=\frac{v \sin \theta}{\sqrt{2}}$
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