Answer
(a) $980KN$
(b) upward
Work Step by Step
(a) We can find the force exerted on the roof as follows:
$F=\Delta PA$
$\implies F=[\frac{1}{2}\rho(v_{top}^2-v_{in}^2)]A$
We plug in the known values to obtain:
$F=\frac{1}{2}(1.29Kg/m^3)[(47.7m/s)^2-0]$
$F=980KN$
(b) We know that the stationary air exerts more pressure than moving air outside, hence the force is exerted on the roof upward.
