Question

A particle is launched from a horizontal plane with speed $u$ and angle of projection $\theta$. The angular velocity of the particle as observed from point of projection at the time of landing will be:

• A. $\frac{g}{2u\cos \theta }$
• B. $\frac{g}{u\cos \theta }$
• C. $\frac{3g}{2u\cos \theta }$
• D. $\frac{2g}{u\cos \theta }$

Answer 1

The correct option is A $\frac{g}{2u\cos \theta }$


When the particle lands, it will hit the ground with same speed of projection i.e., $u$.

With respect to line $\text{OP}$,

$v_{parallel}=v_x=u\cos \theta$

$v_{perpendicular}=v_y=u\sin \theta$

So, angular velocity w.r.t point of projection,

$\omega =\frac{v_{perpendicular}}{r}$

$\Rightarrow \omega =\frac{u\sin \theta }{OP}=\frac{u\sin \theta }{R}$

Here, $R=\frac{u^2\sin 2\theta }{g}$

$\Rightarrow \omega =\frac{gu\sin \theta }{u^2\times 2\sin \theta \cos \theta }$

$\therefore \omega =\frac{g}{2u\cos \theta }$

Hence, option $\left(a\right)$ is correct.

Why this question ? Tip: In problems asking for angular velocity about some point, always figure out the perpendicular component of velocity and distance along the line joining both observer point and particle.