1. The problem statement, all variables and given/known data A greasy spud is launched off of the top of an infinitely large frictionless wedge. The angle that the wedge slopes down at is 30 degrees. Find the launch angle that maximizes the horizontal range of the spud. 2. Relevant equations x = x0 + vxt + (1/2)at2 3. The attempt at a solution I began by rotating the axes so that the x-axis would be parallel with the wedge. Then I divided gravity into x and y components. ax = sin(30)mg ay = cos(30)mg Then I found the total time the spud is in the air t = 2(sinθv0)/(cos(30)mg) The horizontal range of the spud can be written as x = cosθv0t + (1/2)(sinθmg)t2 Substituting for t yields x = ((2sinθv02)/(cos(30))(cosθ + (sin2θ)/(cos(30)) I then took the derivative of cos(θ) + sin2(θ))/(cos(30) with respect to θ and got 2cosθsinθ/cos(30) - sinθ setting this equal to zero and solving for theta yielded 64.3 degrees. Then, subtracting the angle through which I rotated the axes gives 34.3 degrees.