BITSAT 2011 Paper

The velocity of the car is given as, $v_{c}=30\,\text{ms}^{-1}$ And the velocity of ball with respect to the car is given as, $v_{bc}=20\,\text{ms}^{-1}$ Now we need to find the velocity of the ball with respect to the car with its components. $v_{bc}=20 \cos 60^{\circ} i+20 \sin 60^{\circ} j$ $\Rightarrow v_{bc}=20 \times \frac{1}{2} i+20 \times \frac{\sqrt{3}}{2} j$ $\therefore v_{bc}=10 i+10 \sqrt{3} j$ Now, we know that, $v_{bc}=v_{b}-v_{c}$ $\Rightarrow v_{b}=v_{bc}+v_{c}$ $\Rightarrow v_{b}=10 i+10 \sqrt{3} j+30 i$ $\therefore v_{b}=40 i+10 \sqrt{3} j$ Now we need to find the angle of projection. $\tan \theta = \frac{10 \sqrt{3}}{40}$ $\therefore \theta = \tan^{-1}\left(\frac{\sqrt{3}}{4}\right)$ Therefore, the required value of the angle of projection is $\tan^{-1}\left(\frac{\sqrt{3}}{4}\right)$. Hence, option (B), i.e. $\tan^{-1}\left(\frac{\sqrt{3}}{4}\right)$ is the correct choice of the given question.