The position of a projectile launched from the origin at $t = 0$ is given by $\vec{r} = (40\hat{i} + 50\hat{j})\,m$ at $t = 2\,s$. If the projectile was launched at an angle $\theta$ from the horizontal,then $\theta$ is (take $g = 10\,m/s^2$)
- A$\tan^{-1}(\frac{2}{3})$
- B$\tan^{-1}(\frac{3}{2})$
- C$\tan^{-1}(\frac{7}{4})$
- D$\tan^{-1}(\frac{4}{5})$
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