A projectile is thrown with velocity v at an angle θ with horizontal. When projectile is at a heigh

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3-2.Motion in Plane

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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

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

C

$\frac{v \sin \theta}{\sqrt{2}}$

D

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

Solution

(c)

$v_B^2=v^2 \sin ^2 \theta-\frac{2 g}{2}\left(\frac{u^2 \sin ^2 \theta}{2 g}\right)$

$v_B^2=\frac{v^2 \sin ^2 \theta}{2}$

$v_B=\frac{v \sin \theta}{\sqrt{2}}$

Standard 11

Physics

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(AIPMT-2012)