Question

The vertical component of the velocity of projectile is:

• A. $3v\sin \theta$
• B. $v\sin \theta$
• C. $\frac{v\sin \theta }{\sqrt{2}}$
• D. $\frac{v\sin \theta }{\sqrt{3}}$

Answer 1

The correct option is C $\frac{v\sin \theta }{\sqrt{2}}$

Initial velocity $=V$ angle of projectile $=a$

Vertical component of initial velocity $=V\sin \theta$

at maximum height $V_f=0$

$V^2-{\mu }^2=2as$

$0-{\left(V\sin \theta \right)}^2=-2gh$

$h=\frac{{\left(V\sin \theta \right)}^2}{2g}$

$h^1=h^1/2h^1=\frac{{\left(V\sin \theta \right)}^2}{4g}$

time taken to reach half of maximum height.

$h^1=\mu t+\frac{1}{2}{at}^2$

$\frac{V^2{\sin }^2\theta }{4g}=V\sin \theta t+\frac{1}{2}(-10)t^2$

$\Rightarrow {200t}^2-40r\sin \theta t+V^2{\sin }^2\theta =0$

$t=\frac{(\sqrt{2}\pm 1)V\sin \theta }{10\sqrt{2}}$

Vertical component of velocity

$V_y=\mu +at$

$=V\sin \theta -\frac{g\times (\sqrt{2}-1)V\sin \theta }{10\sqrt{2}}$

$V_y=\frac{V\sin \theta }{\sqrt{2}}$