Question

The front of a garage is a square 15 ft on each side with a triangular roof above the square. The height of the triangular roof is 10.6 ft. To the nearest hundred, how much force is exerted by an 80 mi/h wind blowing directly against the front of the garage? Use the formula F = 0.004Av2.

Answer 1

see the picture in the attached figure to better understand the problem

we know that
Area of the square=b*h----->15²----> 225 ft²

Area of the triangular roof=b*h/2-----> 15*10.6/2----< 79.5 ft²

The total area of the front of the garage is= 225 + 79.5----> 304.5 ft²

Find the force of the wind against the front of the garage
Use the formula for force
F = 0.004*A*v²

Substitute 304.5 for A and 80 for v
so
F ≠ 0.004*(304.5)*(80)²------> F=7795.2------> F=7800 lb

the answer is
An 80 mi/h wind exerts a force of about 7800 lb against the front of the garage 

Answer 2

Answer: 7800 lb of force is exerted by  an 80 mi/hr wind blowing directly against the front of the garage.

Step-by-step explanation:

Since we have given that

Side of square = 15 ft

Triangular roof is above the square:

So, Base of triangle = 15 ft

Height of roof = 10.6 ft

velocity of wind (v) = 80 miles/ hr

So, Area of square is given by

$Side\times Side=15\times 15=225\ ft^2$

Similarly,

Area of triangular roof is given by

$\dfrac{1}{2}\times base\times height\\\\=0.5\times 15\times 10.6\\\\=79.5\ ft^2$

So, Total area would be  225+79.5=304.5 ft².

And we have to use " $F=0.004Av^2$ ".

So, we will put A = 304.5 ft² .

So, it becomes,

$F=0.004\times 304.5\times (80)^2\\\\F=7795.2\ lb\\\\F\approx 7800\ lb$

Hence,7800 lb of force is exerted by  an 80 mi/hr wind blowing directly against the front of the garage.