Answer:
We can think the water stream as a solid object that is fired.
The distance between the fireperson and the building is 50m. (i consider that the position of the fireperson is our position = 0)
The angle is 30 above the horizontal. (yo wrote 300, but this has no sense because 300° implies that he is pointing to the ground).
The initial speed of the stream is 40m/s.
First, using the fact that:
x = R*cos(θ)
y = R*sin(θ)
in this case R = 40m/s and θ = 30°
We can use the above relation to find the components of the velocity:
Vx = 40m/s*cos(30°) = 34.64m/s
Vy = 20m/s.
First step:
We want to find the time needed to the stream to hit the buildin.
The horizontal speed is 34.64m/s and the distance to the wall is 50m
So we want that:
34.64m/s*t = 50m
t = 50m/(34.64m/s) = 1.44 seconds.
Now we need to calculate the height of the stream at t = 1.44s
Second step:
The only force acting on the water is the gravitational one, so the acceleration of the stream is:
a(t) = -g.
g = -9.8m/s^2
For the speed, we integrate over time and we get:
v(t) = -g*t + v0
where v0 is the initial speed: v0 = 20m/s.
The velocity equation is:
v(t) = -g*t + 20m/s.
For the position, we integrate again over time:
p(t) = -(1/2)*g*t^2 + 20m/s*t + p0
p0 is the initial height of the stream, this data is not known.
Now, the height at the time t = 1.44s is
p(1.44s) = -5.9m/s^2*(1.44s)^2 + 20m/s*1.44s + po
= 16.57m + p0
So the height at wich the stream hits the building is 16.57 meters above the initial height of the fire hose.
so if we double the charge on 
and double the distance to 
we plug these into the equation to find
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if I can remember my physics correctly.