= Solution
For steady <incompressible flow> of constant density without a body force, the <Euler equation> and mass conservation give
$$
\rho u_j\partial_j u_i=-\partial_i p,\qquad \partial_j u_j=0,
\qquad \partial_j(p\delta_{ij}+\rho u_i u_j)=0.
$$
Apply the <divergence theorem> componentwise to a fluid volume. With $\mathbf n$ its outward normal, the integral <momentum flux> balance is
$$
\boxed{\int_{\partial V}\left(p\mathbf n+\rho\mathbf u(\mathbf u\cdot\mathbf n)\right)dS=\mathbf0.}
$$
If a solid bounds part of the volume, retain its pressure contribution; that contribution supplies the force on the fluid. This derivation includes pressure as well as transported momentum.
Choose the axis in the incoming jet direction. The printed diagram measures $\alpha$ between that axis and the downstream sheet: it is the cone's half-angle. Along a streamline, the steady Euler equation integrates to the <Bernoulli equation> $p/\rho+|\mathbf u|^2/2=\text{constant}$. At the incoming free jet and the outgoing free sheet the pressure is atmospheric. With gravity neglected, the outgoing speed is therefore $u$. This argument uses far free surfaces; it does not assert atmospheric pressure on the wetted sphere.
At radial distance $r$, the sheet's circumference is $2\pi r\sin\alpha$. Its cross-section normal to its velocity is this circumference times its small thickness $d$. <Conservation of mass> gives $Au=(2\pi r\sin\alpha)du$, hence
$$
\boxed{d=\frac{A}{2\pi r\sin\alpha}.}
$$
For the axial momentum balance, transverse contributions cancel by axial symmetry. The outgoing axial momentum rate is $\rho Au^2\cos\alpha$ and the incoming rate is $\rho Au^2$. Uniform atmospheric pressure cancels on the closed control surface when expressed as gauge pressure. The sphere's force on the fluid is therefore $\rho Au^2(\cos\alpha-1)$ in the axis direction. By action and reaction, \b[the fluid force on the sphere is downstream], with magnitude
$$
\boxed{F=\rho Au^2(1-\cos\alpha).}
$$
The downstream direction in the original diagram fixes this sign and the factor $1-\cos\alpha$. If one calls the full cone opening its vertex angle, that full opening is $2\alpha$; using $\alpha$ as the full opening would conflict with the printed thickness formula.
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