Question

A particle is projected at an angle of ${30}^{\circ }$ with respect to the horizontal with speed $20m/s$. The angle between velocity vector and position vector at $t=1s$ is :

• A. $co{s}^{-1}\left(2\sqrt{\frac{3}{13}}\right)$
• B. $si{n}^{-1}\left(2\sqrt{\frac{3}{13}}\right)$
• C. $co{s}^{-1}\left(2\sqrt{\frac{13}{3}}\right)$
• D. $si{n}^{-1}\left(2\sqrt{\frac{13}{3}}\right)$

Answer 1

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

Given, $\theta ={30}^o$,

Initial speed, $u=20m/s$

$v_x=ucos\theta =10\sqrt{3}\hat{i}$

$v_y=u_y+a_{y}t=10-gt=0$

$\therefore \overline{v}=10\sqrt{3}\hat{i}$, $|\overline{v}|=10\sqrt{3}$

$\overline{v}\cdot \overline{r}=\left(10\sqrt{3}\hat{i}\right)\cdot (10\sqrt{3}\hat{i}+5\hat{j})=300$

$\overline{v}\cdot \overline{r}=|\overline{v}||\overline{r}|cos\theta$

$\Rightarrow cos\theta =\frac{\overline{v}\cdot \overline{r}}{|\overline{v}||\overline{r}|}=\frac{300}{10\sqrt{3}\sqrt{325}}$

$\Rightarrow \theta =co{s}^{-1}\left(2\sqrt{\frac{3}{13}}\right)$