Question

A body is projected with initial velocity of $(8\hat{i}+6\hat{j})\text{m/s}$ from origin, find angular velocity of body wrt to an axis passing through origin and perpendicular to the x -y plane when it lands. (Here $\hat{i},\hat{j}$ are unit vectors corresponding to X, Y axis respectively)
Take $g=10{\text{m/s}}^2$

• A. $\frac{5}{8}rad/s$
• B. $\frac{8}{5}rad/s$
• C. $2.4rad/s$
• D. $4.5rad/s$

Answer 1

The correct option is A $\frac{5}{8}rad/s$

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).

Hence the vertical component of velocity when particle lands is $6\text{m/s}$ downward direction or $-6\hat{j}$.

Angular velocity $\omega =\frac{|v_{OA\perp }|}{R}$

Given $\overrightarrow{u}=8i+6j$
$\Rightarrow u_x=8,u_y=6$ and
we know that range of projectile
$\Rightarrow R=\frac{2u_x{u}_y}{g}=\frac{2\times 8\times 6}{10}$,
$\Rightarrow |v_{OA\perp }|=6$

$\omega =\frac{|v_{OA\perp }|}{R}=\frac{6}{R}=\frac{6\times 10}{2\times 8\times 6}=\frac{5}{8}\text{rad/s}$

Hence option A is the correct answer.