Answer
$P_1=1.47 \times 10^5~Pa$
Work Step by Step
$P1+\rho gy_1+{1\over 2}\rho v_1^2=P_2+\rho gy_2+{1\over 2}\rho v_2^2$
Since the main has a much larger diameter than the hose, $v_1=0$ and $v_2=0$ because the final velocity at the top of the stream will be 0. There is no pressure at the top of the stream so $P_2=0$. We will consider the main as $y_1=0$. Therefore, we get the following equation:
$P_1=\rho g y_2$
$P_1=(1000~kg/m^3)(9.8~m/s^2)(15~m)=1.47 \times 10^5~Pa$
