Answer
(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.
Work Step by Step
(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.
