The radius of a cone is proportional to height. In this case, it looks like the constant of proportionality is (4 ft/(10 ft)) = 0.4, so the equation for the volume is

v = (π/3)(0.4h)^2*h = .16π/3*h^3

The rate of change of volume with respect to time is

dv/dt = (.16π/3)*3*h^2 = .16π*h^2*dh/dt

Solving for dh/dt, we get

(dv/dt)/(.16π*h^2) = dh/dt

Using the given values for dv/dt and h, we get

(-10 ft^3/min)/(.16π*(5 ft)^2) = dh/dt

-5/(2π) ft/min = dh/dt

The water level is

v = (π/3)(0.4h)^2*h = .16π/3*h^3

The rate of change of volume with respect to time is

dv/dt = (.16π/3)*3*h^2 = .16π*h^2*dh/dt

Solving for dh/dt, we get

(dv/dt)/(.16π*h^2) = dh/dt

Using the given values for dv/dt and h, we get

(-10 ft^3/min)/(.16π*(5 ft)^2) = dh/dt

-5/(2π) ft/min = dh/dt

The water level is

**dropping at the rate of 5/(2π) ≈ 0.795775 ft/min**.