First of all: Welcome Ahamad!
I'm not gonna give a complete answer, but some hints how to arrive at the result. As you may know, the force $F$ exerted on the water in the pipe is $F=\frac{dp}{dt}$. This means that the force acting on the water surface in the pipe is the infinitesimal change in the momentum in an infinitesimal amount of time. The pressure on the water becomes $\frac{F}{{\frac1 4}\pi d^2}$ (the force divided by the area enclosed by the pipe's circumference). Let's assume all the momentum is conversed in pressure (exerted on the surface and thus on the bottom of the pipe).
To calculate $\frac{dp}{dt}$ we can calculate the change in momentum of $q$ liter water (weighing about $q(kg)$) in one second (which is the same as $\frac{dp}{dt}$ because of the constant stream) and considering the $q(kg)$ as a point mass when reaching the surface (of course the distance of $20(m)$ is in reality diminished if $q(l)$ has fallen on the water in the pipe, but in this way we can calculate $\frac{dp}{dt}$).
We can use the equation $s=\frac{1}{2}10t^2$ to calculate the time it takes if $s=20(m)$, after which we can use $v=10t$ to calculate the velocity of the $q(kg)$, and thus the momentum change in one second, which is the same as $\frac{dp}{dt}$.
The more water has entered the pipe, the less the pressure exerted by the falling water on the surface (because it falls a smaller distance), until it's zero when the pipe is full, in which case there is only the weight of the water contributing to the pressure.
So initially, the only pressure comes from the falling water (and the little water initially there, which is negligible) after which the pressure of the water on the bottom of the pipe increases by the increasing amount of water in the pipe, and the pressure caused by the falling water decreases because it falls less distance. A more thorough calculation will reveal the total pressure if the water level rises. Try!
