## College Physics (4th Edition)

The plane's airspeed is $272~m/s$
The pressure difference between the moving air and the stationary air is $\frac{1}{2}\rho_a~v^2$ This pressure difference is equal to the gauge pressure in the manometer, which is $\rho_m~g~h$ We can equate the expressions for the pressure difference to find the airspeed $v$: $\frac{1}{2}\rho_a~v^2 = \rho_m~g~h$ $v^2 = \frac{2~\rho_m~g~h}{\rho_a}$ $v = \sqrt{\frac{2~\rho_m~g~h}{\rho_a}}$ $v = \sqrt{\frac{(2)~(13600~kg/m^3)~(9.80~m/s^2)(0.25~m)}{0.90~kg/m^3}}$ $v = 272~m/s$ The plane's airspeed is $272~m/s$.