Afire hose held near the ground shoots water at a speed of 6.5 m/s. at what angle(s) should the nozle point in order that the water land 2.0 m away, why are there two different angles?

in the diagram, the hose is held straight up at theta degrees going at 6.5 m/s and with water falling down.. the distance between initial and final water displacement is 2m.. a fire hose held near the ground shoots water at a speed of 6.5 m/s. at what angle(s) should the nozle point in order that the water land 2.0 m away, why are there two different angles?

 The trajectory (parabola) for a "particle" of water follows from: 

h(t) = v(0) sin(angle) t - 1/2 g t^2 

x(t) = v(0) cos(angle) t 

From the second equation t = x / (v(0) cos(angle) ) 
and substitution in the first to eliminate t gives the parabola h(x): 

h(x) = x tan(angle) - (g/(2v(0)^2 cos(angle)^2) x^2 

h(x) = 0 is solved by x=0 (where the trajectory starts) and at 

x = 2 v(0)^2 tan(angle) cos^2(angle) / g 

x = 2 v(0)^2 sin(angle) cos(angle) / g 

x = v(0)^2 sin(2 angle) / g [[ using sin(2 ) = 2 sin(s) cos(s) ]] 

For the first question (the 2.0m distance question) you have therefore to solve for angle the equation 

2.0= 2*(6.5)^2 sin(2 angle) / 9.81 

sin(2 angle) = 0.23

angle = 1/2 * arcsin(0.23) = 6.65 degrees 

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