Let the conical container on the right be empty first. Let the radius of the water surface in the container be r and the height be h at time t, then r=h/2 4 = h/8 and r=h/2. The volume of water V = π r 2h/3 = π h 3/ 12.

DV/DH = π h 2/4, DV/DT = DV/DH× DH/DT. If DV/DT = 4 and H = 5, DH/DT = 4/(25π), so when the water depth is 5 meters, the water surface rises at a speed of 4/(25π) (m/s).