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Initial Angle = 52.5 degrees from the horizontal
Initial velocity = 20 m/s

Net Placement:
vx = v*cos(theta)
vx = 12.175 m/s
vy = v*sin(theta)
vy = 15.867 m/s
time = (change in x)/(v*cos(theta))
time = (change in x)/12.175 s
a = -9.8 m/s^2
delta x = vx*t
t = (delta x)/vx
delta y = vy*t+1/2*a*t^2
delta y = 15.867m/s*(delta x)/(12.175m/s)-4.9*((delta x)^2)/(148.2382m^2/s^2)
delta y = 1.3032*delta x - .03306*(delta x)^2
0 = 1.3032*delta x - .03306*(delta x)^2
delta x = 0m and delta x = 39.4252m
The net should be 39.4252m away at the same initial hight from which the human cannonball is fired.

Hoop placement:
delta x = (39.4252m)/2 = 19.7126 m
t = (delta x)/vx
t = 19.7126m/(12.175m/s)
t = 3.2381 s
delta y = 15.867m/s*t-4.9*t^2
delta y = 15.867m/s*3.2381s-4.9m/s^2*(3.2381s)^2
delta y = 12.8451 m
The center of the hoop should be 19.7126 meters away in the x direction and 12.8451 meters away in the y direction