Explanation:
\(\vec u = \left( {40\,\,\cos 30^\circ \,\hat i + 40\,\,\sin 30^\circ \,\hat j} \right){\mkern 1mu} \,m/s\)
\( = (20\sqrt 3 \hat i + 20\hat j)m{\rm{/}}s\)
\(\vec g = - 10\hat j,\;\;\;{\mkern 1mu} {\kern 1pt} t = 2\,{\rm{sec}}\)
Now, \(\vec{v}=\vec{u}+\vec{g} t=(20 \sqrt{3} \hat{i}+20 \hat{j})+(-10 \hat{j}) \times 2\)
\( \Rightarrow \vec v = 20\sqrt 3 \hat i\;m{\rm{/}}s\)
\(\therefore \vec v = 20\sqrt 3 \;m{\rm{/}}s\).