A fire hose held near the ground shoots water at a speed of 6.5 m/s. At what angle should the nozzle point in order that the water will land 2.5 m away?

Question

anonymous · Answer

As the horizontal component of the velocity '6.5cos theta' is constant everywhere of this trajectory, just use this eqn, S=6.5 cos 'theta'.  The influence of g is not applicable here because g only acts vertically. Here, 'theta' is the angle at what the nozzle as well as the initial path of water should be pointed. S is the horizontal distance given.
Hope you got the answer.

anonymous · Answer

First of all you should find elapsed time t. For that, you can use this Vfin=Vin-g*t  
Here t is required time  that water molecules reach top of the flight. So you should multiply it to 2.

Horizontal component of the velocity is 

Vx=6.5*Cos(theta) 

and the vertical is 

Vy=6.5*Sin(theta) 

Then you can use this 

$$R=\frac{ v ^{2}*\sin2\theta }{ g }$$

anonymous · Answer

So can you help me plug the numbers into that equation to get the answer? I'm still confused