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Q.
At a height 0.4 m from the ground, the velocity of a projectile in vector form is:$\vec{v}=\left(6\hat{i}+2\hat{j}\right)$ m/s. The angle of projection is:
Motion in a Plane
Solution:
$v^{2} =u^{2} -2gh \,\,$ or $\,\,u^{2} = v^2 +2gh$
or $u^{2}_{x} +u^{2}_{y} =v^{2}_{x} +v^{2}_{y} +2gh$
As $v_{x} =u_{x}; u^{2}_{y} =v^{2}_{y} + 2 gh$
or $u^{2}_{y} =(2)^{2} +2 \times 10\times 0.4=12$
$u_{y} =\sqrt{12} =2\sqrt{3} \,m/s$
$u_{x} =v_{x} =6 \,m/s$
tan $\theta=\frac{u_{y}}{u_{x}} =\frac{2\sqrt{3}}{6}=\frac{1}{\sqrt{3}}$
$\theta=30^{\circ}$