Answer
See a detailed answer below.
Work Step by Step
a) Since the water exits into the air at a constant speed, the pressure of the water at this point is atmospheric pressure of 1 atm.
So,
$$P_{\rm hose}=P_a=\color{red}{\bf 1.013\times 10^5}\;\rm Pa$$
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b) As we see in the figure below, we are going to use Bernoulli’s equation at points A and B where $A$ is the point below the vertical column of water and $B$ is at the open hose. The vertical distance between them is 4 cm as shown in the original figure given in the problem.
Thus,
$$ P_A+\frac{1}{2}\rho v_A^2+\rho gy_A=P_B+\frac{1}{2}\rho v_B^2+\rho gy_B$$
So,
$$P_A-P_B= \frac{1}{2}\rho v_B^2+\rho gy_B-(\frac{1}{2}\rho v_A^2+\rho gy_A)$$
$$P_A-P_B= \frac{1}{2}\rho (v_B^2-v_A^2)+\rho g(y_B- y_A) $$
Noting that $P_A=P_a+h\rho g$ and $P_B=P_a$
$$\color{red}{\bf\not} P_a+h\rho g-\color{red}{\bf\not} P_a = \frac{1}{2}\rho (v_B^2-v_A^2)+\rho g(y_B- y_A) $$
$$ h\color{red}{\bf\not} \rho g = \frac{1}{2}\color{red}{\bf\not} \rho (v_B^2-v_A^2)+\color{red}{\bf\not} \rho g(y_B- y_A) $$
Hence,
$$h= \frac{1}{2g} (v_B^2-v_A^2)+ (y_B- y_A) \tag 1$$
Now we need to find $v_A$ at the left part of the pipe which we can find by applying the continuity equation.
$$A_Av_A=A_Bv_B$$
We can see that $A_A=10\;\rm cm^2$ while $A_B=5\;\rm cm^2$ which means that $A_A=2A_B$
$$(2\color{red}{\bf\not} A_B)v_A=\color{red}{\bf\not} A_Bv_B$$
Hence,
$$v_A=\frac{1}{2}v_B$$
Plug into (1) and then plugging the given;
$$h= \frac{1}{2g} \left(v_B^2-\left[\frac{1}{2}v_B\right]^2\right)+ (y_B- y_A) $$
$$h= \frac{v_B^2}{2g} \left(1- \frac{1}{4} \right)+ (y_B- y_A) $$
$$h= \frac{3v_B^2}{8g} + (y_B- y_A) $$
$$h= \frac{3(4)^2}{8(9.8)} + 4 =\color{red}{\bf 4.61}\;\rm m$$
