Solve Kinematics Fireman Question: 28.7m, 49.5°, 39.6m/s

[pkossak]

A fireman, 28.7 m away from a burning building, directs a stream of water from a ground level fire hose at an angle of 49.5° above the horizontal. If the speed is 39.6 m/s, at what height will the stream of water hit the building?

Got this right on the first test, and can't figure it out again right now for some reason?!

Anyway, I figured 39.6cos49.5 should be Vxo, and a = Vxo^2/28.7 m
That gives me Vxo = 25.72 m/s, a = 23.05 m/s^2
I have a suspicion that if something is wrong, it has to do with how I calculated a - am I right?

Anyway, I plugged that into x = xo + Vxo*t + 0.5*a*t^2

Then I found t, and plugged all of the numbers I have into
y = yo + vyo*t - 0.5*g*t^2

of course, vyo = 39.6sin49.5 = 30.112 m/s

I keep getting the wrong answer

Any suggestions? Thanks a ton

 

[Doc Al]

The acceleration is known: It's 9.8 m/s^2 downward.

What you need to calculate is the time it takes for the water to travel the horizontal distance to the building. Then use that in your equation for vertical postion to find the height.

 

[pkossak]

in trying to find the time that water is traveling horizontally, don't I need a horizontal acceleration?

 

[whozum]

Nope, there is no acceleration horizontally. It has a velocity component in the horizontal direciton which is what you need to use to find the time it takes.

 

[pkossak]

got it, thanks a lot.