1. Statement of the problem. Find the center of pressure of a vertical rectangular gate of width $b=3$ m and height $h=7$ m whose top edge is at depth $y_{top}=2.5$ m below the free surface, with density $\rho=1000$ kg/m^3 and $g=9.81$ m/s^2. 2. Formulas and important rules. The hydrostatic pressure at depth $y$ is given by the formula $$p(y)=\rho g y$$. The total hydrostatic force on the plane is $$F=\rho g\int_A y\,dA$$. For a vertical plane the center of pressure depth $y_p$ measured from the free surface is $$y_p = y_c + \frac{I_{xc}}{y_c A}$$ where $y_c$ is the centroid depth, $A$ is the area, and $I_{xc}$ is the second moment of area about the centroidal horizontal axis. Important rule: density and $g$ factor out of the ratio used to find $y_p$, so $y_p$ depends only on geometry and depth to centroid. 3. Intermediate work and evaluation. Centroid depth from free surface: $$y_c = y_{top} + \frac{h}{2} = 2.5 + \frac{7}{2} = 6.0$$. Area of the rectangular gate: $$A = b h = 3 \times 7 = 21$$. Second moment of area about centroidal horizontal axis: $$I_{xc} = \frac{b h^3}{12} = \frac{3 \times 7^3}{12} = 85.75$$. Compute the corrective term: $$\frac{I_{xc}}{y_c A} = \frac{85.75}{6.0 \times 21} = 0.6805555556$$. Compute center of pressure depth from free surface: $$y_p = y_c + \frac{I_{xc}}{y_c A} = 6.0 + 0.6805555556 = 6.6805555556$$. Compute center of pressure depth measured from the top edge of the gate: $$y_{p,top} = y_p - y_{top} = 6.6805555556 - 2.5 = 4.1805555556$$. 4. Final answer. The center of pressure is at depth $y_p\approx 6.6806$ m measured from the free surface. Equivalently, it is $4.1806$ m below the top edge of the gate.