Question

A particle is launched from a horizontal plane with speed $10\text{m/s}$ and angle of projection ${60}^{\circ }$. The angular velocity of the particle as observed from the point of projection at the time of landing will be:

• A. $1\text{rad/sec}$
• B. $2\text{rad/sec}$
• C. $3\text{rad/sec}$
• D. $4\text{rad/sec}$

Answer 1

The correct option is A $1\text{rad/sec}$


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

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

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

Given, $g=10{\text{m/s}}^2,\theta ={60}^{\circ },u=10{\text{m/s}}^2$

$\Rightarrow \omega =\frac{10}{2\times 10\cos {60}^{\circ }}=1\text{rad/s}$

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