Physics: Principles with Applications (7th Edition)

(a) The water gauge pressure at the house is $7.0\times 10^5~Pa$ (b) The water could reach a height of 71.5 m.
(a) We can find the vertical height of the 75-meter pipe. $\frac{h_p}{75~m} = sin(61^{\circ})$ $h_p = (75~m)~sin(61^{\circ})$ $h_p = 65.5~m$ The total change in height from the top of the tank to the bottom of the pipe is 71.5 m We can find the gauge pressure at the bottom of the pipe. $P_G = \rho~g~h$ $P_G = (1.00\times 10^3~kg/m^3)(9.80~m/s^2)(71.5~m)$ $P_G = 7.0\times 10^5~Pa$ The water gauge pressure at the house is $7.0\times 10^5~Pa$. (b) We can use Bernoulli's equation to find the initial speed when the water leaves the pipe. $P_2+\frac{1}{2}\rho v_2^2 = P_1+\frac{1}{2}\rho v_1^2$ $P_0+\frac{1}{2}\rho v_2^2 = P_0 + P_G + \frac{1}{2}\rho (0)^2$ $\frac{1}{2}\rho v_2^2 = \rho~g~h$ $v_2^2 = 2~g~h$ $v_2 = \sqrt{2gh}$ $v_2 = \sqrt{(2)(9.80~m/s^2)(71.5~m)}$ $v_2 = 37.44~m/s$ The speed when the water comes out of the pipe is 37.44 m/s. We can use kinematics to find the maximum height H. $v_f^2-v_0^2 = 2gH$ $H = \frac{v_f^2-v_0^2}{2g}$ $H = \frac{(0)^2-(37.44~m/s)^2}{(2)(-9.80~m/s^2)}$ $H = 71.5~m$ The water could reach a height of 71.5 m. Note that this is the height of the top of the water tank.