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Solution: Position $(x_{s},y_{s})$ of the stagnation point for the rankine half-body.

From the expression for the potential, it is obvous that $y_{s} = 0$. So, we need to find $x_{s}$. First, we obtain the $x$ component of the velocity vector.


\begin{displaymath}u = \frac{\partial \phi }{\partial x} = U + \frac{m}{2\pi }\frac{x}{x^2 +y^2} \notag
\end{displaymath}

Now we set $u = 0$ and solve for $x$. We already know that $y_{s} = 0$, so we set $y = 0$ in the expression for $u$ to obtain $x_{s}$.


\begin{displaymath}U + \frac{m}{2\pi x} = 0 \Rightarrow x = - \frac{m}{2\pi U} \notag
\end{displaymath}  

Therefore,


\begin{displaymath}u = 0 \mbox{\ at\ } x_{s} = - \frac{m}{2\pi U}. \notag
\end{displaymath}