Water is leaking out of a reservoir so that the height of the water, h (in metres), satisfies the differential equation: dh/dt = - k h^(1/2), where time, t, is measured in hours and k is a particular case constant. Determine height of the water (in metres) after 25 hours, solving differential equation of the system numerically, using MATLAB. Use the following particular case data: initial water level was h0 = 4 m, and after 5 hrs it dropped to 3 m.

Given the differential equation:

dh/dt = -k√(h)

We can separate variables and integrate both sides to solve it:

Separating variables:

∫√(h) dh = -k ∫dt

Integrating both sides:

(2/3)h^(3/2) = -kt + C

Given that when t=0, h=4, we can find C:

(2/3)(4)^(3/2) = 0 + C
C = 32/3

The equation becomes:

(2/3)h^(3/2) = -kt + 32/3

Given that after 5 hours h=3 m, we can find k:

(2/3)(3)^(3/2) = -5k + 32/3
18 = -5k + 32/3
(-5k) = -14/3
k = 14/15

Now we can find the height of the water after 25 hours:

(2/3)h^(3/2) = - (14/15)t + 32/3

Plugging in t=25:

(2/3)h^(3/2) = - (14/15)(25) + 32/3
(2/3)h^(3/2) = -70 + 32/3
h^(3/2) = (-105 + 64) / 2
h^(3/2) = 41/2
h = (41/2)^(2/3)
h = 4.19 m

Therefore, the height of the water after 25 hours would be 4.19 metres.

\boxed{h = 4.19 \text{ m}}