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
    $\frac{v \sin \theta}{\sqrt{3}}$

Answer

Correct option: C.
$\frac{v \sin \theta}{\sqrt{2}}$
c
(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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