The pipe acts as a Pitot tube, which measures fluid velocity by converting kinetic energy to potential energy. The open end pointing upstream captures the stagnation pressure. Using Bernoulli's principle for a streamline starting in the free river and ending at the stagnation point inside the pipe:

$\mathrm{P}+\frac{1}{2}{\mathrm{\rho v}}^2+\mathrm{\rho gh}={\mathrm{P}}_0+\mathrm{\rho gH}$

Assuming the pipe and river surface are at the same level, the static pressure P is atmospheric.

The height h is the same, so they cancel. The stagnation pressure P₀ is also atmospheric.

Solving for the height H in the vertical arm gives $\mathrm{H}=\frac{{\mathrm{v}}^2}{2\mathrm{g}}$.

Final Answer:$\mathrm{H}=\frac{{\left(5\right)}^2}{2\times 9.8}\approx 1.28\text{ m}$