Question

A particle is projected at an angle of ${30}^{\circ }$ w.r.t horizontal with speed $20m/s.$ Take the point of projection as the origin of the coordinate axes and find the velocity vector of the particle after 1 s (in $m/s$).

• A. $5\sqrt{3}\hat{i}+5\hat{j}$
• B. $5\hat{j}$
• C. $5\hat{i}+10\sqrt{3}\hat{j}$
• D. $10\sqrt{3}\hat{i}$

Answer 1

The correct option is D $10\sqrt{3}\hat{i}$

Since the horizontal component of velocity always remains constant
$v_x=u_x=u\cos {30}^{\circ }=20\times \frac{\sqrt{3}}{2}\Rightarrow \overrightarrow{v_x}=10\sqrt{3}\hat{i}m/s{v}_y=u_y-gt{v}_y=u\sin {30}^{\circ }-gt\Rightarrow v_y=10-10=0\therefore \overrightarrow{v}=10\sqrt{3}\hat{i}m/s$