The horizontal range of a projectile is $4 \sqrt{3}$ times of its maximum height. The angle of projection will be

The position of a projectile launched from the origin at $$t=0$$ is given $$\mathbf{r}=(40 \hat{\mathbf{i}}+50 \hat{\mathbf{j}}) \mathrm{m}$$ at $$t=2 \mathrm{~s}$$. If the projectile was launded at an angle $$\theta$$ from the horizontal, then $$\theta$$ is (take, $$g=10 \mathrm{~m} / \mathrm{s}^2$$ )