1. The problem statement, all variables and given/known data At what projection angle will the range of a projectile equal its maximum height? 2. Relevant equations Big 5 vf2= vi2 + 2ad 3. The attempt at a solution So I attempted to solve this question by randomly subbing in equations into each other. If i draw out a diagram to solve for the angle, I would need the x and y components. Let V equal the total velocity. viy= Vsinθ Plugging into vfy2= viy2 + 2aydy , I get vf2= V2sin2θ + 2aydy Now I isolate for d, and end up with dy= (V2sin2θ)/2ay Ok, after I got the displacement vertically, since I am trying to find the point where they both are at equal magnitude, I now solve for displacement horizontally. Now using the range formula, I can determine the horizontal displacement. dx= (V2sin2θ)/g (g=ay) With these two formulas, I substitute and... (V2sin2θ)/2ay = (V2sin2θ)/ay sin2θ = aysin2θ sin2θ = ay(2sinθcosθ) sinθ = 2aycosθ sinθ/cosθ = 2ay tanθ = 2(-9.8) θ = tan-1(-19.6) θ = 87° However... this is not the answer that the answer key states. The answer shown is 76°, and I do not know where I have gone wrong, or if I am doing this right. Sorry if it was a mess and confusing, thanks for the help!