What is the average force exerted by the water on the wall when it rebounds?

[longball4153]

Homework Statement

A horizontal stream of water from a fire-fighter's hose hits a vertical wall. The firefighter knows that 16.9 kg of water are emitted from the hose per second, and that the water has a speed of 45.5 m/s. At the instant when it hits the wall, the direction of the velocity vector of the water is 25.0 degrees BELOW the horizontal. Calculate the average force exerted by the water on the wall, assuming that the vertical component of the velocity vector of the water is unchanged but the horizontal component of the velocity vector is reversed when the water rebounds from the wall.

V_i=45.5 m/s
m= 16.9 kg/s
ϴ= 25 Degrees BELOW the horizontal

Homework Equations

P=mv
v²=v_o²+2ax

The Attempt at a Solution

P_o=mv_o
p_o=16.9kg/s*45.5m/s
P_o=768.95

Now I have the initial momentum. From here I am stuck and not sure where to go next. I know I need to somehow incorporate the angle given and calculate a final velocity. But then how would I take that and find an average force? Hopefully someone can explain/ show me the set up from here. Thanks!

 

[Bacat]

An average force is defined as the sum of the forces applied divided by the period of time. For example, if you have force a for 1 second and force b for 2 seconds, then the average force is:

\frac{1s*F_a + 2s*F_b}{3s} where s is seconds.

Does that help?

 

[longball4153]

I understand that, however, how do I relate that to this particular problem?

 

[cmgroff]

I am doing this exact problem right now. I have found the momentum of the water and the x component of the velocity. Average force is the change in momentum/change in time, but how can you find time using the information in the problem?