- #1
pkossak
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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
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