1. The problem statement, all variables and given/known data 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. 2. Relevant 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 3. 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.