Question

A particle is projected at an angle ${60}^{\circ }$ with the horizontal with a speed $10m/s$.
Find angular velocity when particle reaches X-axis, about an axis passing through origin ang along Z-axis.
Take $g=10{\text{m/s}}^2$

• A. $1rad/s$
• B. $2rad/s$
• C. $3rad/s$
• D. $4rad/s$

Answer 1

The correct option is A $1rad/s$

Here in the ground to ground projectile, when the particle lands the vertical component of velocity will have same magnitude, only the direction changes to downwards and the horizontal component remains same (both magnitude and direction).
$v_{OA\perp }=u\sin {60}^{\circ }$


Angular velocity:
$\omega =\frac{v_{OA\perp }}{R}=\frac{u\sin {60}^{\circ }}{R}$

Here Range of projectile
$R=\frac{u^2\sin 2\theta }{g}=\frac{100\times \frac{\sqrt{3}}{2}}{10}=5\sqrt{3}$

$\omega =\frac{u\sin {60}^{\circ }}{R}=\frac{10\times \frac{\sqrt{3}}{2}}{5\sqrt{3}}=1rad/s$