Hi

I know that I am answering this question 5 years after your posting this question.I am a new member.I think I was able to solve this question and got the correct answer of 7.84 m/s,but with a lot of assumptions! So I am posting this answer to get yours and others' opinion about my solution.

My assumptions:-

1) Imagine a rectangle and imagine pulling it from its center,raising a pyramid.I assume the roof to be of this pyramidal shape so that,the exposed surface area of this pyramid is same as that of the base.(This is not the case with all pyramids)

2) Pressure inside the house is the same as the atmospheric pressure.

3) Water collected in the pipe joining the floor drainage and the pipe M is stagnant(at rest)

Solution:

Let the point in M directly below the floor drainage be called point 1),the point where a downspout begins as point 3) and the point at the floor drainage as point 2).

Consider a y axis such that y1=0,y2=h2,y3=h1

Between 1) and 2)

P1=P2 + Rg(y2-y1) (water is at rest)

P1= Pa + Rgh2

Since 3) is exposed to the atmosphere,P3=Pa

Therefore,P1-P3=Rgh2

Applying Bernoulli's equation between 1) and 3):-

P1+0.5R(v1^2)+Rgy1=P3+0.5R(v3^2)+Rgy3

Therefore,

P1-P3=Rgh1-0.5R(v1^2)

P1-P3=Rgh2

Therefore,

Rgh2=Rgh1-0.5R(v1^2)

This gives v1=13.85929291m/s

Volume flow rate=A1v1=0.039186227

Continuity:-

Volume flow rate=Rainfall volume rate rate

Now,a point on the roof receives a volume (vdt).da in a time dt where v=speed of rain and da=area of point.Therefore,volume rate=vda

Integrating,flow rate=lbv where l and b anre length and breadth.

lbv=A1v1

Solving the above would give v=2.177 x (10^-5) m/s=7.84 cm/h

Please tell me if I am correct or not