= Solution
For a steady uniform jet, the mass flow rate is $\dot m=\rho Au$. The transported <momentum flux> and <kinetic energy flux of a uniform jet> are
$$
\boxed{\dot p=\rho Au^2,\qquad\dot K=\frac12\rho Au^3.}
$$
Incompressibility gives piston speed $v_p=Au/S$. Equating piston power $Fv_p$ to the jet's kinetic power gives
$$
\boxed{F=\frac12\rho Su^2.}
$$
This ideal result neglects losses and the kinetic energy of fluid moving slowly inside the large cylinder. Retaining an inlet speed correction would multiply the pressure expression by $1-(A/S)^2$.
A horizontal impact occurs at the projectile apex. Write $u_x=u\cos\theta$ and $u_y=u\sin\theta$. At its height h, $u_y^2=2gh$ and the flight time is $t=u_y/g$, so
$$
u_y=\sqrt{2gh},\qquad u_x=w\sqrt{\frac g{2h}},\qquad
\boxed{u=9.90\,\mathrm{m\,s^{-1}},\quad\theta=45^\circ,\quad F=4.91\,\mathrm N.}
$$
Here $g=9.81\,\mathrm{m\,s^{-2}}$ and $u_x=u_y=7.00\,\mathrm{m\,s^{-1}}$. The <apex-target condition for a ballistic jet> fixes both launch speed and angle.
If the target stops the horizontal momentum of the arriving water, the force is
$$
F_{\rm hit}=\dot m\,u_x=\rho A u u_x=0.277\,\mathrm N,
\qquad \mu m_tg=0.147\,\mathrm N.
$$
\b[This exceeds the limiting friction, so the target can be dislodged] in the ideal stopping-impact model. The <jet-area adjustment during ballistic slowing> matters: the nozzle area sets $\dot m$; replacing u by $u_x$ in that mass-flow factor would incorrectly assume unchanged jet area during deceleration. Reflection or a specified deflection would change the transferred momentum. Air drag, breakup and splash losses are excluded from the ballistic estimate.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2011/paper-1-ia2011-jet-apex.png]
{title=Ballistic jet apex}
{source=paper-1-ia2011-jet-apex.py}
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