Solution: Position $(x_{s},y_{s},z_{s})$ of the stagnation point for the 3D rankine half-body.

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

$$u = \frac{\partial \phi }{\partial x} = U + \frac{m}{4\pi }\frac{x}{(x^2 +y^2+z^{2})^{3/2}} \notag$$

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

$$U + \frac{m}{4\pi}\frac{x}{\vert x\vert^{3}} = 0 \Rightarrow x = - \sqrt{\frac{m}{4\pi U}} \notag$$

Therefore,

$$u = 0 \mbox{\ at\ } x_{s} = - \sqrt{\frac{m}{4\pi U}}. \notag$$