The dimensional analysis of the given equation is solved as
[tex]\begin{gathered} \frac{1}{2}mv^2=\frac{1}{2}mv^2_0+\sqrt[]{mgh} \\ \lbrack M^1L^2T^{-2}\rbrack=\lbrack ML^2T^{-2}\rbrack+\lbrack M^1L^2T^{-2}\rbrack^{\frac{1}{2}} \\ \lbrack M^1L^2T^{-2}\rbrack=\lbrack M^{\frac{3}{2}}L^3T^{-3}\rbrack \end{gathered}[/tex]As the left-hand side of the dimension is not equal to the right-hand side of the dimension.
Hence, it is dimensionally incorrect.
