- #1
vetgirl1990
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Homework Statement
The mountain rescue officers trigger the avalanche by firing a gun at a distant mountain slope.
The muzzle velcoity of the gun is 225m/s. The horizontal and vertical distance from the gun to the slope is
x=1,072m and y=538m respectively. Find the lowest angle between the barrel of the gun and the horizontal plane.
Homework Equations
rf = ri + vt + 1/2at2
Dividing each into the horizontal and vertical components:
- Horizontal component: xf = xi + vxt + 1/2axt2
- Vertical component: yf = yi + vyt + 1/2ayt2
Vxi = VicosΘ
Vyi = VisinΘ
The Attempt at a Solution
I started off by manipulating the equations in terms of time (t), since I don't have that either.
In the x direction, the bullet is under constant velocity, so a=0; xf = 1072m; xi = 0
xf = xi + vxt + 1/2axt2
1072 = 225cosΘ t
(1) t = 1072 / 225cosΘ
In the y direction, the bullet is under constant acceleration due to gravity, ay = -g = -9.8m/s2
yf = yi + vyt + 1/2ayt2
528 = 225sinΘ t - 4.9t2
(2) 538 = t(225sinΘ - 4.9t)
Then I substituted equation (1) into (2)...
538 = [1072/225cosΘ] [225sinΘ - 4.9(1072/225cosΘ)
538 = 1072sinΘ/cosΘ - 111.22966/cos2Θ
and this is where the trig substitutions start getting messy. It gets quite long, and I ended up with 1.211249 = sin2Θ, which is impossible.
So either I am approaching the question wrong, or I am not manipulating my trig properly.
Any help or direction would be greatly appreciated.