The "power available in the wind" of velocity \(V\) through an area \(A\) is

\[ \dot{W}=\frac{1}{2} ho A V^{3} \]

where \(ho\) is the air density \(\left(0.075 \mathrm{lbm} / \mathrm{ft}^{3}\right)\). For an \(18-\mathrm{mph}\) wind, find the wind area \(A\) that will supply a power of \(4 \mathrm{hp}\).