1. The problem statement, all variables and given/known data My problem consists of two parts, the first being how far from the top of a cliff will a projectile (fired from a cannon) land with a velocity of 120 ms^-1, fired horizontally from cliff height of 150m. The second part is where I am struggling, the cannon is now aimed 45° above horizontal, how far will the projectile travel now? Known data for part 1. Δy = -150 u(y) = 0 ms^-1 a(y) = -9.8 ms^-2 u(x) = 120 ms^-1 2. Relevant equations Δy = u(y)*t + 1/2*a(y)*t^2 Δx = u(x)*t x any y component equations. u(x) = u*cos(θ) u(y) = u*sin(θ) 3. The attempt at a solution Part 1 I found quite simple so hopefully this is correct.. Δy = u(y)*t + 1/2*a(y)*t^2 -150 = 0*t + 1/2(-9.8)*t^2 (substitute above data) -150 = -4.9*t^2 t^2 = -150/-4.9 t = 5.53 seconds Δx = u(x)*t Δx = 120*5.53 Δx = 663.9 m Part 2. u(x) = 120*cos(45) u(x) = 84.85 ms^-1 (2 d.p.) u(y) = 120*sin(45) u(y) = 84.85 ms^-1 This is as far I got (finding the x and y components of the initial velocity). My issue is that I am not sure how to account for the fact that the cannon is elevated 150 m along with the angle change from horizontal to 45°. Any help will be greatly appreciated.