MCQ
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 ...........
- A$v \sin \theta \times 3$
- B$\frac{v \sin \theta}{3}$
- ✓$\frac{v \sin \theta}{\sqrt{2}}$
- D$\frac{v \sin \theta}{\sqrt{3}}$




