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 angle between velocity vector and position at $t=1s$

• A. ${\cos }^{-1}\left(\sqrt{\frac{3}{13}}\right)$
• B. ${\cos }^{-1}\left(2\sqrt{\frac{3}{13}}\right)$
• C. ${\cos }^{-1}\left(4\sqrt{\frac{3}{13}}\right)$
• D. ${30}^{\circ }$

Answer 1

The correct option is B ${\cos }^{-1}\left(2\sqrt{\frac{3}{13}}\right)$

Since the horizontal component of velocity always remains constant
$v_x=u_x=u\cos {30}^{\circ }=20\times \frac{\sqrt{3}}{2}\Rightarrow \overrightarrow{v_x}=10\sqrt{3}\hat{i}m/s{v}_y=u_y-gt{v}_y=u\sin {30}^{\circ }-gt\Rightarrow v_y=10-10=0\therefore \overrightarrow{v}=10\sqrt{3}\hat{i}m/s\overrightarrow{v}.\overrightarrow{r}=\left(10\sqrt{3}\hat{i}\right)(10\sqrt{3}\hat{i}+5\hat{j})=300m^2/s$
Now,
$\overrightarrow{v}.\overrightarrow{r}=|\overrightarrow{v}||\overrightarrow{r}|\cos \alpha \Rightarrow \cos \alpha =\frac{\overrightarrow{v}.\overrightarrow{r}}{|\overrightarrow{v}||\overrightarrow{r}|}\Rightarrow \cos \alpha =\frac{300}{10\sqrt{3}\sqrt{325}}\Rightarrow \alpha ={\cos }^{-1}\left(2\sqrt{\frac{3}{13}}\right)$