Q. 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
Motion in a Plane
Solution:
$v_{B}^{2}=v^{2} \sin ^{2} \theta-\frac{2 g}{2}\left(\frac{u^{2} \sin ^{2} \theta}{2 g}\right)$
$v_{B}^{2}=\frac{v^{2} \sin ^{2} \theta}{2}$
$v_{B}=\frac{v \sin \theta}{\sqrt{2}}$
