A projectile is fired in such a way that its horizontal range is equal to three times its maximum height. What is the angle of projection? If I let A = maximum height (or h), than horizontal range (or R) = 3A. I suppose I can also substitute 1 for h and 3 for R. There is some relationship between the horizontal velocity and the initial vertical velocity that allows for this feat to be accomplished. The angle must be of about medium amount, about 40-60 degrees? Equations include the components of vf = vi + at, components of rf = ri + vit + .5a(t^2), equations for range and height: h = ((vi^2)((sinθ)^2))/(2g), where g = -9.8 m/s^2 time it takes for particle to reach max height: th = (vi(sinθ))/g R = vxi2th = ((vi^2)(sin2θ))/g θ = angle of projection, but all of these equations require some knowledge of the initial velocity, which I don't have. And the equations involving that include time, which I also do not have. I think I'm not getting something important from the information already given. The point at the beginning is (0,0), while at maximum height, the coordinates are (1.5,1) or (3A, A). So, now I think that I can use the rules of parabolic equations to figure this out. But, I don't think they want me to figure this problem out like that.