A fire hose held near the ground shoots water at a speed of 6.5M/S . At what angle should the nozzle point in order that the water land 2.0M away?

Question

anonymous · Answer

is this related with porjectile

anonymous · Answer

using special formula because it,s symetrical 
R=v^2sin 2(theta)/g
so from here you can find (theta)

anonymous · Answer

$$\theta = \frac{ 1 }{2}\left( \sin^{-1} \frac{ 2*9.8 }{ \left( 6.5 \right)2 } \right)$$

anonymous · Answer

$$\theta = \sin^{-1} (x*g/v^{2})/2 $$
theta is the angle of the nozzle
x is horizontal displacement (2 meters)
g is the gravity (9.8 meter/sec*sec)
v is the initial velocity (6.5 meter/sec)
theta=arcsin(2*9.8/6.5**2)/2
theta=0.24119926291439242 radians
theta*180/pi = 13.819 degrees !
almost horizontal !